Fractional equation reducible to second degree

Statement

Solve the fractional equation, stating the domain conditions: $\frac{3}{x-1}+\frac{2}{x+2}=1.$

Solution

Domain: $x\neq 1$ and $x\neq -2$. Multiplying by $(x-1)(x+2)$: $3(x+2)+2(x-1)=(x-1)(x+2)\Rightarrow 5x+4=x^2+x-2\Rightarrow x^2-4x-6=0.$ $x=\dfrac{4\pm\sqrt{16+24}}{2}=2\pm\sqrt{10}$, i.e. $x\approx 5{.}162$ or $x\approx -1{.}162$. Both admissible.

$\boxed{x=2\pm\sqrt{10}\approx 5{.}162;\ -1{.}162}$