Solve by Factoring x^2-2x-10=5

$x^{2} - 2 x - 10 = 5$

Step 1

Subtract $5$ from both sides of the equation.

$x^{2} - 2 x - 10 - 5 = 0$

Step 2

Subtract $5$ from $- 10$ .

$x^{2} - 2 x - 15 = 0$

Step 3

Factor $x^{2} - 2 x - 15$ 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 $- 15$ and whose sum is $- 2$ .

$- 5, 3$

Write the factored form using these integers.

$(x - 5) (x + 3) = 0$

Step 4

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

$x - 5 = 0$

$x + 3 = 0$

Step 5

Set $x - 5$ equal to $0$ and solve for $x$ .

Set $x - 5$ equal to $0$ .

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$

Step 6

Set $x + 3$ equal to $0$ and solve for $x$ .

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 7

The final solution is all the values that make $(x - 5) (x + 3) = 0$ true.

$x = 5, - 3$

Solve by Factoring x^2-2x-10=5

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