## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$d=\left\{ -5,8 \right\}$
Expressing in the form $ax^2+bx+c=0,$ the given equation is equivalent to \begin{array}{l}\require{cancel} d(d-3)=40 \\\\ d^2-3d=40 \\\\ d^2-3d-40=0 .\end{array} Using the factoring of trinomials in the form $x^2+bx+c,$ the $\text{ equation }$ \begin{array}{l}\require{cancel} d^2-3d-40=0 \end{array} has $c= -40$ and $b= -3 .$ The two numbers with a product of $c$ and a sum of $b$ are $\left\{ -8,5 \right\}.$ Using these two numbers, the $\text{ equation }$ above is equivalent to \begin{array}{l}\require{cancel} (d-8)(d+5)=0 .\end{array} Equating each factor to zero (Zero Product Property), and then isolating the variable, the solutions are $d=\left\{ -5,8 \right\} .$