# What are the x-intercept(s) of the graph of y + 12 = x^2 + x?

Jun 11, 2017

See a solution process below:

#### Explanation:

To find the $x$-intercepts we need to set $y$ to $0$ and solve for $x$:

$y + 12 = {x}^{2} + x$ becomes:

$0 + 12 = {x}^{2} + x$

$12 - \textcolor{red}{12} = {x}^{2} + x - \textcolor{red}{12}$

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

$0 = \left(x + 4\right) \left(x - 3\right)$

Solution 1)

$x + 4 = 0$

$x + 4 - \textcolor{red}{4} = 0 - \textcolor{red}{4}$

$x + 0 = - 4$

$x = - 4$

Solution 2)

$x - 3 = 0$

$x - 3 + \textcolor{red}{3} = 0 + \textcolor{red}{3}$

$x - 0 = 3$

$x = 3$

The $x$-intercepts are: $- 4$ and $3$

Or

$\left(- 4 , 0\right)$ and $\left(3 , 0\right)$