Solve $\dfrac{2}{x+1} - 3 = \dfrac{1}{2x-5}$. Answer ...

1. State the restrictions on $x$ and multiply by the common denominator

Since the denominators cannot be zero,
$x+1\ne 0,\qquad 2x-5\ne 0$
so
$x\ne -1,\qquad x\ne \frac52.$
Multiply
$\frac{2}{x+1}-3=\frac{1}{2x-5}$
by $(x+1)(2x-5)$:
$2(2x-5)-3(x+1)(2x-5)=x+1.$

2. Expand and simplify to form a quadratic equation

$4x-10-3(2x^2-3x-5)=x+1$
$4x-10-6x^2+9x+15=x+1$
$-6x^2+13x+5=x+1$
$-6x^2+12x+4=0$
$3x^2-6x-2=0$

3. Solve the quadratic equation

Solving
$3x^2-6x-2=0$
using the quadratic formula,
$x=\frac{-(-6)\pm\sqrt{(-6)^2-4(3)(-2)}}{2(3)}$
$x=\frac{6\pm\sqrt{36+24}}{6}$
$x=\frac{6\pm\sqrt{60}}{6}$
$x=\frac{3\pm\sqrt{15}}{3}.$

4. Reject no invalid roots and give the solutions

The solutions
$x=\frac{3+\sqrt{15}}{3}\quad\text{or}\quad x=\frac{3-\sqrt{15}}{3}$
are not $-1$ or $\frac52$, so both are valid.