Answer
5 and 2
Work Step by Step
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.
