Real World Examples of Quadratic Equations
An example of a Quadratic Equation:
Quadratic equations pop up in many real world situations!
Here we have collected some examples for you, and solve each using different methods:
 Factoring Quadratics
 Completing the Square
 Graphing Quadratic Equations
 The Quadratic Formula
 Online Quadratic Equation Solver
Each example follows three general stages:
 Take the real world description and make some equations
 Solve!
 Use your common sense to interpret the results
Balls, Arrows, Missiles and Stones
If you throw a ball (or shoot an arrow, fire a missile or throw a stone) it will go up into the air, slowing down as it goes, then come down again ...
... and a Quadratic Equation tells you where it will be!
Example: Throwing a Ball
A ball is thrown straight up, from 3 m above the ground, with a velocity of 14 m/s. When does it hit the ground?
Ignoring air resistance, we can work out its height by adding up these three things:
(Note: t is time in seconds)
The height starts at 3 m:  3  
It travels upwards at 14 meters per second (14 m/s):  14t  
Gravity pulls it down, changing its speed by about 5 m/s per second (5 m/s^{2}):  −5t^{2}  
(Note for the enthusiastic: the 5t^{2} is simplified from (½)at^{2} with a=9.81 m/s^{2}) 
Add them up and the height h at any time t is:
h = 3 + 14t − 5t^{2}
And the ball will hit the ground when the height is zero:
3 + 14t − 5t^{2} = 0
Which is a Quadratic Equation ! In "Standard Form" it looks like:
−5t^{2} + 14t + 3 = 0
Let us solve it ...
There are many ways to solve it, here we will use the factoring method:
Multiply all terms by −1 to make it easier: 5t^{2} − 14t − 3 = 0  
Now our job is to factor it. We will use the "Find two numbers that
multiply to give a×c, and add to a×c = −15, and b = −14. The factors of −15 are: −15, −5, −3, −1, 1, 3, 5, 15 By trying a few combinations we find that −15 and 1 work 

Rewrite middle with −15 and 1:  5t^{2} − 15t + t − 3 = 0  
Factor first two and last two:  5t(t − 3) + 1(t − 3) = 0  
Common Factor is (t − 3):  (5t + 1)(t − 3) = 0  
And the two solutions are:  5t + 1 = 0 or t − 3 = 0  
t = −0.2 or t = 3 
The "t = −0.2" is a negative time, impossible in our case.
The "t = 3" is the answer we want:
The ball hits the ground after 3 seconds!
Here is the graph of the Parabola h = −5t^{2} + 14t + 3
It shows you the height of the ball vs time
Some interesting points:
(0,3) When t=0 (at the start) the ball is at 3 m
(−0.2,0) says that −0.2 seconds BEFORE we threw the ball it was at ground level. This never happened! So our common sense says to ignore it.
(3,0) says that at 3 seconds the ball is at ground level.
Also notice that the ball goes nearly 13 meters high.
Note for the enthusiastic: You can find exactly where the top point is!
The method is explained in Graphing Quadratic Equations, and has two steps:
Find where (along the horizontal axis) the top occurs using −b/2a:
 t = −b/2a = −(−14)/(2 × 5) = 14/10 = 1.4 seconds
Then find the height using that value (1.4)
 h = −5t^{2} + 14t + 3 = −5(1.4)^{2} + 14 × 1.4 + 3 = 12.8 meters
So the ball reaches the highest point of 12.8 meters after 1.4 seconds.
Example: New Sports BikeYou have designed a new style of sports bicycle! Now you want to make lots of them and sell them for profit. 
Your costs are going to be:
 $700,000 for manufacturing setup costs, advertising, etc
 $110 to make each bike
Based on similar bikes, you can expect sales to follow this "Demand Curve":
 Unit Sales = 70,000 − 200P
Where "P" is the price.
For example, if you set the price:
 at $0, you would just give away 70,000 bikes
 at $350, you would not sell any bikes at all.
 at $300 you might sell 70,000 − 200×300 = 10,000 bikes
So ... what is the best price? And how many should you make?
Let us make some equations!
How many you sell depends on price, so use "P" for Price as the variable
 Unit Sales = 70,000 − 200P
 Sales in Dollars = Units × Price = (70,000 − 200P) × P = 70,000P − 200P^{2}
 Costs = 700,000 + 110 x (70,000 − 200P) = 700,000 + 7,700,000 − 22,000P = 8,400,000 − 22,000P
 Profit = SalesCosts = 70,000P − 200P^{2} − (8,400,000 − 22,000P) = −200P^{2} + 92,000P − 8,400,000
Profit = −200P^{2} + 92,000P − 8,400,000
Yes, a Quadratic Equation. Let us solve this one by Completing the Square.
Solve: −200P^{2} + 92,000P − 8,400,000 = 0
Step 1 Divide all terms by 200
Step 2 Move the number term to the right side of the equation:
Step 3 Complete the square on the left side of the equation and balance this by adding the same number to the right side of the equation:
(b/2)^{2} = (−460/2)^{2} = (−230)^{2} = 52900
Step 4 Take the square root on both sides of the equation:
Step 5 Subtract (230) from both sides (in other words, add 230):
What does that tell us? It says that the profit is ZERO when the Price is $126 or $334
But we want to know the maximum profit, don't we?
It is exactly half way inbetween! At $230
And here is the graph:
Profit = −200P^{2} + 92,000P − 8,400,000
The best sale price is $230, and you can expect:
 Unit Sales = 70,000 − 200 x 230 = 24,000
 Sales in Dollars = $230 x 24,000 = $5,520,000
 Costs = 700,000 + $110 x 24,000 = $3,340,000
 Profit = $5,520,000 − $3,340,000 = $2,180,000
A very profitable venture.
Example: Small Steel Frame
Your company is going to make frames as part of a new product they are launching.
The frame will be cut out of a piece of steel, and to keep the weight down, the final area should be 28 cm^{2}
The inside of the frame has to be 11 cm by 6 cm
What should the width x of the metal be?
Area of steel before cutting:
Area of steel after cutting out the 11 × 6 middle:
Let us solve this one graphically!
Here is the graph of 4x^{2} + 34x :
The desired area of 28 is shown as a horizontal line.
The area equals 28 cm^{2} when:
x is about −9.3 or 0.8
The negative value of x make no sense, so the answer is:
x = 0.8 cm (approx.)
Example: River Cruise
A 3 hour river cruise goes 15 km upstream and then back again. The river has a current of 2 km an hour. What is the boat's speed and how long was the upstream journey?
There are two speeds to think about: the speed the boat makes in the water, and the speed relative to the land:
 Let x = the boat's speed in the water (km/h)
 Let v = the speed relative to the land (km/h)
Because the river flows downstream at 2 km/h:
 when going upstream, v = x−2 (its speed is reduced by 2 km/h)
 when going downstream, v = x+2 (its speed is increased by 2 km/h)
We can turn those speeds into times using:
time = distance / speed
(to travel 8 km at 4 km/h takes 8/4 = 2 hours, right?)
And we know the total time is 3 hours:
total time = time upstream + time downstream = 3 hours
Put all that together:
total time = 15/(x−2) + 15/(x+2) = 3 hours
Now we use our algebra skills to solve for "x".
First, get rid of the fractions by multiplying through by (x2)(x+2):
3(x2)(x+2) = 15(x+2) + 15(x2)
Expand everything:
3(x^{2}−4) = 15x+30 + 15x−30
Bring everything to the left and simplify:
3x^{2} − 30x − 12 = 0
It is a Quadratic Equation! Let us solve it using the Quadratic Formula:
Where a, b and c are
from the
Quadratic Equation in "Standard Form": ax^{2} + bx + c = 0
Solve 3x^{2}  30x  12 = 0
Coefficients are:  a = 3, b = −30 and c = −12  
Quadratic Formula:  x = [ −b ± √(b^{2}−4ac) ] / 2a  
Put in a, b and c:  x = [ −(−30) ± √((−30)^{2}−4×3×(−12)) ] / (2×3)  
Solve:  x = [ 30 ± √(900+144) ] / 6  
x = [ 30 ± √(1044) ] / 6  
x = ( 30 ± 32.31 ) / 6  
x = −0.39 or 10.39 
Answer: x = −0.39 or 10.39 (to 2 decimal places)
x =−0.39 makes no sense for this real world question, but x = 10.39 is just perfect!
Answer: Boat's Speed = 10.39 km/h (to 2 decimal places)
And hence the upstream journey = 15 / (10.39−2) = 1.79 hours = 1 hour 47min
And the downstream journey = 15 / (10.39+2) = 1.21 hours = 1 hour 13min
Example: Resistors In Parallel
Two resistors are in parallel, like in this diagram:
The total resistance has been measured at 2 Ohms, and one of the resistors is known to be 3 ohms more than the other.
What are the values of the two resistors?
The formula to work out total resistance "R_{T}" is:
\frac{1}{R_{T}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}
In this case, we have R_{T} = 2 and R_{2} = R_{1} + 3
\frac{1}{2} = \frac{1}{R_{1}} + \frac{1}{R_{1}+3}
Now, let us set about solving this:
Get rid of the fractions by first multiplying all terms by 2R_{1}(R_{1} + 3): 
\frac{2R_{1}(R_{1}+3)}{2} = \frac{2R_{1}(R_{1}+3)}{R_{1}} + \frac{2R_{1}(R_{1}+3)}{R_{1}+3} 

Then simplify:  R_{1}(R_{1} + 3) = 2(R_{1} + 3) + 2R_{1}  
Expand to:  R_{1}^{2} + 3R_{1} = 2R_{1} + 6 + 2R_{1}  
Bring all terms to the left:  R_{1}^{2} + 3R_{1} − 2R_{1} − 6 − 2R_{1} = 0  
Simplify:  R_{1}^{2} − R_{1} − 6 = 0 
Yes! A Quadratic Equation !
Let us solve it using our Quadratic Equation Solver.
 Enter 1, −1 and −6
 And you should get the answers −2 and 3
R_{1} cannot be negative, so R_{1} = 3 Ohms is the answer.
The two resistors are 3 ohms and 6 ohms.
Others
Quadratic Equations are useful in many other areas:
For a parabolic mirror, a reflecting telescope or a satellite dish, the shape is defined by a quadratic equation.
Quadratic equations are also needed when studying lenses and curved mirrors.
And many questions involving time, distance and speed need quadratic equations.