Solve for x x^2+2x-48=0

$x^{2} + 2 x - 48 = 0$

Step 1

Factor $x^{2} + 2 x - 48$ 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 $- 48$ and whose sum is $2$ .

$- 6, 8$

Write the factored form using these integers.

$(x - 6) (x + 8) = 0$

Step 2

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

$x - 6 = 0$

$x + 8 = 0$

Step 3

Set $x - 6$ equal to $0$ and solve for $x$ .

Set $x - 6$ equal to $0$ .

$x - 6 = 0$

Add $6$ to both sides of the equation.

$x = 6$

Step 4

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 5

The final solution is all the values that make $(x - 6) (x + 8) = 0$ true.

$x = 6, - 8$

Solve for x x^2+2x-48=0

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