Answer
The solutions are $m=11$ and $m=4$.
Work Step by Step
$m^{2}+10=15m-34\implies m^{2}-15m+44=0$
To factor the polynomial $m^{2}+bm+c$ to $(m+p)$ and $(m+q)$, we need to have $p+q=b$ and $pq=c$.
In this case, $b=-15$ and $c=44$.
$\implies p=-11$ and $q=-4$.
Then, $m^{2}-15m+44=(m-11)(m-4)$
$m^{2}-15m+44=0\implies (m-11)(m-4)=0$
Using zero-product property, we have
$m-11=0$ or $m-4=0$
$\implies m=11$ or $m=4$
The solutions are $m=11$ and $m=4$.
