Answer
$x=-2$ and $x=-3$ satisfy the given conditions.
Work Step by Step
$y_{1}=x-3$ $,$ $y_{2}=x+8$ and $y_{1}y_{2}=-30$
If $y_{1}y_{2}$ is equal to $-30$, then $(x-3)(x+8)$ is also equal to $-30$.
$(x-3)(x+8)=-30$
Evaluate the product on the left:
$x^{2}+8x-3x-24=-30$
Take $-30$ to the left side and simplify:
$x^{2}+8x-3x-24+30=0$
$x^{2}+5x+6=0$
Solve by factoring:
$(x+2)(x+3)=0$
Set both factors equal to $0$ and solve each individual equation for $x$:
$x+2=0$
$x=-2$
$x+3=0$
$x=-3$
$x=-2$ and $x=-3$ satisfy the given conditions.
