Question

What are the solutions to the equation?

x2 + 6x = 40

x = −10 and x = 4
x = −8 and x = 5
x = −5 and x = 8
x = −4 and x = 10

Answer 1

To solve the equation \( x^2 + 6x = 40 \), we start by rearranging it into standard form:

\[
x^2 + 6x - 40 = 0
\]

Next, we can apply the quadratic formula, which is given by:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

In this case, \( a = 1 \), \( b = 6 \), and \( c = -40 \).

First, we calculate the discriminant:

\[
b^2 - 4ac = 6^2 - 4(1)(-40) = 36 + 160 = 196
\]

Next, we calculate the square root of the discriminant:

\[
\sqrt{196} = 14
\]

Now, we can substitute back into the quadratic formula:

\[
x = \frac{-6 \pm 14}{2(1)}
\]

This gives us two possible solutions:

1. Using the plus sign:

\[
x = \frac{-6 + 14}{2} = \frac{8}{2} = 4
\]

2. Using the minus sign:

\[
x = \frac{-6 - 14}{2} = \frac{-20}{2} = -10
\]

Thus, the solutions to the equation \( x^2 + 6x = 40 \) are:

\[
x = -10 \quad \text{and} \quad x = 4
\]

The correct answer is:

**x = −10 and x = 4**