# How do you solve y = 3x + 12 and y = x^2 + 2x - 18?

May 2, 2016

The points of intersections are: $\left(6 , 30\right)$ and $\left(- 5 , - 3\right)$

#### Explanation:

We are looking for the intersection of a line and a parabola. In other words, we want the points where if we put the same $x$ into both equations, we get the same $y$, therefore we should set the two $y$'s equal to each other to get the equation:

$3 x + 12 = {x}^{2} + 2 x - 18$

which we can simplify to

$0 = {x}^{2} - x - 30$

Now we can use the quadratic equation to find the roots of our new quadratic where $a = 1$, $b = - 1$, and $c = - 30$

$x = \frac{1 \pm \sqrt{1 + 120}}{2} \implies 6 , - 5$

To find the corresponding $y$ values we can use either equation, but it's simpler to use the line:

$y = 3 x + 12 \implies 30 , - 3$

So the points of intersections are: $\left(6 , 30\right)$ and $\left(- 5 , - 3\right)$