## Algebra 2 (1st Edition)

$x \pm 4$
We have $3x+16=x(x+3)$ ...(1) Re-write as: $3x+16=x^2+3x$ or, $x^2-16 =0 \implies (x-4)(x+4)=0$ or, $x \pm 4$ and $x \ne -3, \dfrac{-16}{3}$ The solutions are not undefined values. So, $x \pm 4$