Answer
$x=-4$ and $x=3$
Work Step by Step
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 $x=3$.
