Solve for x x^2+4x-40=-8

$x^{2} + 4 x - 40 = - 8$

Step 1

Add $8$ to both sides of the equation.

$x^{2} + 4 x - 40 + 8 = 0$

Step 2

Add $- 40$ and $8$ .

$x^{2} + 4 x - 32 = 0$

Step 3

Factor $x^{2} + 4 x - 32$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 32$ and whose sum is $4$ .

$- 4, 8$

Write the factored form using these integers.

$(x - 4) (x + 8) = 0$

Step 4

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 4 = 0$

$x + 8 = 0$

Step 5

Set $x - 4$ equal to $0$ and solve for $x$ .

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

Step 6

Set $x + 8$ equal to $0$ and solve for $x$ .

Set $x + 8$ equal to $0$ .

$x + 8 = 0$

Subtract $8$ from both sides of the equation.

$x = - 8$

Step 7

The final solution is all the values that make $(x - 4) (x + 8) = 0$ true.

$x = 4, - 8$

Solve for x x^2+4x-40=-8

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