Answer
$\{4+\sqrt 5, 4-\sqrt 5 \}$
Work Step by Step
$\frac{5}{x+1} + \frac{x-1}{4} = 2$
$x=-1$ makes the denominator zero, so,
$\frac{5}{x+1} + \frac{x-1}{4} = 2; x\ne -1$
Taking LCD,
$\frac{20+(x+1)(x-1)}{4(x+1)} =2 ; x\ne -1$
$\frac{20+x^{2}-1}{4(x+1)} =2; x\ne -1$
$\frac{x^{2}+19}{4(x+1)} =2; x\ne -1$
$x^{2}+19=8(x+1); x\ne -1$
$x^{2}+19=8x+8; x\ne -1$
$x^{2}+19-8x-8=0; x\ne -1$
$x^{2}-8x+11=0; x\ne -1$
Using quadratic formula,
$x= \frac{-b±\sqrt (b^{2}-4ac)}{2a}$
Substituting $a=1, b=-8, c=11$
$x= \frac{-(-8)±\sqrt ((-8)^{2}-4(1)(11))}{2(1)}; x\ne -1$
$x= \frac{8±\sqrt (64-44)}{2}; x\ne -1$
$x= \frac{8±\sqrt 20}{2}; x\ne -1$
$x= \frac{8±\sqrt 4 \times 5}{2}; x\ne -1$
$x= \frac{8±2\sqrt 5}{2}; x\ne -1$
$x= \frac{2(4±\sqrt 5)}{2}; x\ne -1$
$x=4±\sqrt 5$
$x=4+\sqrt 5$ or $x=4-\sqrt 5$
