Algebra 2 Common Core

$x=4$ and $x=-1$
Subtract $8$ from each side: $$2x^2-6x-8=0$$ Factor out $2$: $$2(x^2-3x-4)=0$$ Factor the trinomial: $$2(x-4)(x+1)=0$$ Use the Zero-Product Property by equating each factor with a variable to zero, then solve each equation to obtain: \begin{align*} x-4&=0 &\text{or}& &x+1=0\\ x&=4 &\text{or}& &x=-1 \end{align*} Thus, the solutions are $x=4$ and $x=-1$.