## Intermediate Algebra: Connecting Concepts through Application

The solution set is $\left\{-4, 1\right\}$.
Simplify the equation by multiplying the binomials and adding $24$ to both sides: \begin{align*} d(d)+d(-4)+7(d)+(7(-4)+24&=-24+24\\ d^2-4d+7d-28+24&=0\\ d^2+3d-4&=0 \end{align*} Factor the trinomial: $$(d+4)(d-1)=0$$ Use the Zero-Product Property by equating each factor to zero. Then, solve each equation to obtain: \begin{align*} d+4&=0 &\text{or}& &d-1=0\\ d&=-4 &\text{or}& &d=1\\ \end{align*} Checking: \begin{align*} (-4+7)(-4-4)&=-24\\ (3)(-8)&=-24\\ -24&\stackrel{\checkmark}=-24 \end{align*} \begin{align*} (1+7)(1-4)&=-24\\ (8)(-3)&=-24\\ -24&\stackrel{\checkmark}=-24 \end{align*} Therefore, the solution set is $\left\{-4, 1\right\}$.