Answer
The solutions are $b=5$ and $b=3$.
Work Step by Step
$b^{2}+5=8b-10\implies b^{2}-8b+15=0$
To factor the polynomial $b^{2}+db+c$ to $(b+p)$ and $(b+q)$, we need to have $p+q=d$ and $pq=c$.
In this case, $d=-8$ and $c=15$.
$\implies p=-5$ and $q=-3$.
Then, $b^{2}-8b+15=(b-5)(b-3)$
$b^{2}-8b+15=0\implies (b-5)(b-3)=0$
Using zero-product property, we have
$b-5=0$ or $b-3=0$
$\implies b=5$ or $b=3$
The solutions are $b=5$ and $b=3$.
