Answer
The solutions are $x=-7$ and $x=2$.
Work Step by Step
To factor the polynomial $x^{2}+bx+c$ to $(x+p)$ and $(x+q)$, we need to have $p+q=b$ and $pq=c$.
In this case, $b=5$ and $c=-14$.
$\implies p=7$ and $q=-2$.
Then, $x^{2}+5x-14=(x+7)(x-2)$
$x^{2}+5x-14=0\implies (x+7)(x-2)=0$
Using zero-product property, we have
$x+7=0$ or $x-2=0$
$\implies x=-7$ or $x=2$
The solutions are $x=-7$ and $x=2$.
