## Algebra 2 Common Core

$x=12 \text{ and } x=-8$
Subtract $100$ from each side: $$x^2-4x-96=0$$ Factor the trinomial: $$(x-12)(x+8)=0$$ Use the Zero-Product Property by equating each factor to zero, then solve each equation to obtain: \begin{align*} x-12&=0 &\text{or}& &x+8=0\\ x&=12 &\text{or}& &x=-8 \end{align*} Thus, the solutions are $x=12 \text{ and } x=-8$.