## Algebra 2 (1st Edition)

We use cross multiplication to solve the equation. Doing this, we find: $$2\left(x^2-2x-3\right)=\left(x-3\right)\cdot \:1 \\ x=3,\:x=-\frac{1}{2}$$ Since 3 is an extraneous solution, $-1/2$ is the actual solution.