## Calculus with Applications (10th Edition)

Set the $RHS$ to 0 by adding 10 to both sides $m(m-7)=-10\qquad.../$+10 $m^{2}-7m+10=0$ To factor the trinomial, we search for factors of +10,\ whose sum is -7$:\qquad$ We find $-5$ and $-2$. Our equation becomes $(m-5)(m-2)=0$ By the zero product principle, $m-5=0$ or $m-2=0$ $m=5$ or $m=2$ The solutions are 5 and 2.