Answer
$x=3$ and $x=-6$ satisfy the given conditions
Work Step by Step
$y_{1}=x-1$ $,$ $y_{2}=x+4$ and $y_{1}y_{2}=14$
If $y_{1}y_{2}$ is equal to $14$, then $(x-1)(x+4)$ is also equal to $14$.
$(x-1)(x+4)=14$
Evaluate the product on the left side:
$x^{2}+4x-x-4=14$
Take $14$ to the left side and simplify:
$x^{2}+4x-x-4-14=0$
$x^{2}+3x-18=0$
Solve by factoring:
$(x-3)(x+6)=0$
Set both factors equal to $0$ and solve each individual equation for $x$:
$x-3=0$
$x=3$
$x+6=0$
$x=-6$
$x=3$ and $x=-6$ satisfy the given conditions