# Let f(x)=x^2+2x-15. Determine the vaules of x for which f(x)=-12?

Feb 7, 2018

$x = \left\{- 3 , 1\right\}$

#### Explanation:

Setting $f \left(x\right) = - 12$ gives us:

$- 12 = {x}^{2} + 2 x - 15$

To solve quadratic equations, you need to set the equation equal to zero. By adding 12 to both sides, we get:

$0 = {x}^{2} + 2 x - 3$

From here, we can factor the quadratic to $0 = \left(x + 3\right) \left(x - 1\right)$

Using the Zero Product Property, we can solve the equation by setting each factor equal to zero and solving for x.

$x + 3 = 0 \to x = - 3$

$x - 1 = 0 \to x = 1$

The two solutions are -3 and 1

Feb 7, 2018

x=-3 and x=1 .

#### Explanation:

Put f(x)=-12

$- 12 = {x}^{2} + 2 x - 15$
${x}^{2} + 2 x - 15 + 12 = 0$
${x}^{2} + 2 x - 3 = 0$

Time to factorize now
${x}^{2} + 3 x - x - 3 = 0$
$x \left(x + 3\right) + \left(- 1\right) \left(x + 3\right) = 0$

take x+3 common
$\left(x + 3\right) \left(x - 1\right) = 0$

x=-3 and x=1 .

Feb 7, 2018

$1$ or $- 3$

#### Explanation:

Since $f \left(x\right) = - 12$, then ${x}^{2} + 2 x - 15 = - 12$. Solve this by factoring:

${x}^{2} + 2 x - 3 = 0$

$\left(x - 1\right) \cdot \left(x + 3\right) = 0$

$x - 1 = 0$

$x + 3 = 0$

$x = 1 , - 3$