## Algebra 2 Common Core

$x=-4$ and $x=3$
Subtract $49$ from each side: $$4x^2+4x-48=0$$ Factor out the common factor $4$ $$4(x^2+x-12)=0$$ Divide $4$ to both sides: $$x^2+x-12=0$$ Factor the trinomial: $$(x+4)(x-3)=0$$ Use the Zero-Product Property by equating each factor to zero, then solve each equation to obtain: \begin{align*} x+4&=0 &\text{or}& &x-3=0\\ x&=-4 &\text{or}& &x=3 \end{align*} Therefore, the solutions are $x=-4$ and $x=3$.