1. Use the Quadratic Formula to solve 3x2 + 23x + 40 = 0.
Lesson 5-8
Additional Examples
Use the Quadratic Formula to solve 3x2 + 23x + 40 = 0.
3x2 + 23x + 40 = 0
a = 3, b = 23, c = 40
Write in standard form.
Find values of a, b, and c.
x =
Write the Quadratic Formula. =
–23 ± – 4(3)(40)
2(3)
Substitute.
–b ± b2 – 4ac 2a
–23 ± 7 6 =
–23 ± 49
–23 ± – 480
Simplify.
= – or – 30 6 16
= –5 or – 8 3

2. (continued) – + 40 0 Check: 3x2 + 23x + 40 = 0 3x2 + 23x + 40 = 0
The Quadratic Formula
Lesson 5-8
Additional Examples
(continued) –
Check: 3x2 + 23x + 40 = 0
3x2 + 23x + 40 = 0
3(–5)2 + 23(–5)
75 –
0 = 0
3 – – 8 3 64
184 2

3. Find the values of a, b, and c.
The Quadratic Formula
Lesson 5-8
Additional Examples
Solve 3x2 + 2x = –4.
3x2 + 2x + 4 = 0
Write in standard form.
a = 3, b = 2, c = 4
Find the values of a, b, and c.
–(2) ± (2)2 – 4(3)(4)
2(3)
y =
Substitute.
–2 ± 4 – 48 6
–2 ± –44 =
–2 ± 2i 11
Simplify.
i 11 3
= – ± 1

4. Find the values of a, b, and c.
The Quadratic Formula
Lesson 5-8
Additional Examples
The longer leg of a right triangle is 1 unit longer than the shorter leg. The hypotenuse is 3 units long. What is the length of the shorter leg?
x2 + (x + 1)2 = 32
x2 + x2 + 2x + 1 = 9
2x2 + 2x – 8 = 0
Write in standard form.
–b ± b – 4ac 2a 2
a = 2, b = 2, c = –8
x =
Find the values of a, b, and c.
Use the Quadratic Formula. =
–(2) ± (2)2 – 4(2)(–8)
2(2)
Substitute for a, b, and c.
–1 ± 17 2 =
Simplify.
17 – 1 2
The length of the shorter leg is units.

5. Is the answer reasonable? Since is a negative number, and a
The Quadratic Formula
Lesson 5-8
Additional Examples
(continued)
– 1 – 17 2
Is the answer reasonable? Since is a negative number, and a
length cannot be negative, that answer is not reasonable.
Since , that answer is reasonable.
17 – 1

6. Determine the type and number of solutions of x2 + 5x + 10 = 0.
The Quadratic Formula
Lesson 5-8
Additional Examples
Determine the type and number of solutions of
x2 + 5x + 10 = 0.
a = 1, b = 5, c = 10 Find the values of a, b, and c.
b2 – 4ac = (5)2 – 4(1)(10) Evaluate the discriminant.
= 25 – 40 Simplify.
= –15
Since the discriminant is negative, x2 + 5x + 10 = 0 has two imaginary solutions.

7. Write the equation in standard form.
The Quadratic Formula
Lesson 5-8
Additional Examples
A player throws a ball up and toward a wall that is 17 ft high. The height h in feet of the ball t seconds after it leaves the player’s hand is modeled by h = –16t t + 6. If the ball makes it to where the wall is, will it go over the wall or hit the wall?
h = –16t t + 6
17 = –16t t + 6
Substitute 17 for h.
0 = –16t t – 11
Write the equation in standard form.
a = –16, b = 25, c = –11
Find the values of a, b, and c.
b2 – 4ac = (25)2 – 4(–16)(–11)
Evaluate the discriminant.
= 625 – 704
= –79
Simplify.
Since the discriminant is negative, the equation 17 = –16t2 + 25t + 6 has no real solution. The ball will hit the wall.