Answer
The solutions are $a=-10$ and $a=5$.
Work Step by Step
$a^{2}+5a-20=30\implies a^{2}+5a-50=0$
To factor the polynomial $a^{2}+ba+c$ to $(a+p)$ and $(a+q)$, we need to have $p+q=b$ and $pq=c$.
In this case, $b=5$ and $c=-50$.
$\implies p=10$ and $q=-5$.
Then, $a^{2}+5a-50=(a+10)(a-5)$
$a^{2}+5a-50=0\implies (a+10)(a-5)=0$
Using zero-product property, we have
$a+10=0$ or $a-5=0$
$\implies a=-10$ or $a=5$
The solutions are $a=-10$ and $a=5$.
