Algebra and Trigonometry 10th Edition

$x=-1, 2$
$f(x)=x^2$ and $g(x)=x+2$ Since we are solving for all the values of $x$ when $f(x) = g(x)$, set the function $f(x)=x^2$ equal to $g(x)=x+2$ and we get the following equation: $x^2=x+2$ Subtract $x+2$ from both sides of the equation so that all the terms are on the left side: $x^2-(x+2)=x+2-(x+2)$ $x^2-x-2=0$ Then, factor the left side of the equation. To do this, we need to find two numbers that multiply to get -2 and add to get -1. Those numbers are 1 and -2. $(x+1)(x-2)=0$ Next, you can see that the above equation holds true when $x+1=0$ or $x-2=0$. Solve for $x$ in each equation individually as shown below: $x+1=0$ Subtract 1 from both sides. $x+1-1=0-1$ $x=-1$ Now for the other equation: $x-2=0$ Add 2 to both sides of the equation: $x-2+2=0+2$ $x=2$ So, the final solution is $x=-1, 2$