Answer
$-3, -\frac{3}{4}$
Work Step by Step
$(\frac{5}{2+x})^{2}+(\frac{5}{2+x})-20=0$
Let $u = (\frac{5}{2+x})$ Equation $(1)$
$u^{2}+u-20=0$
Splitting the middle term into two factors whose product is $-20$ and whose sum is, $1$. The factors are $5$ and $-4$
$u^{2}+5u-4u-20 = 0$
$u(u+5)-4(u+5)=0$
$(u+5)(u-4)=0$
$u = -5$ or $u = 4$
Substituting $u$ values in Equation $(1)$ to find $x$
$u = \frac{5}{2+x}$
$-5 = \frac{5}{2+x}$
$-10-5x = 5$
$-5x = 5+10$
$-5x = 15$
$x = -3$
$u = \frac{5}{2+x}$
$4 = \frac{5}{2+x}$
$8+4x = 5$
$4x = 5-8$
$4x = -3$
$x = -\frac{3}{4}$
