Answer
$x=\dfrac{1}{2}\pm\dfrac{\sqrt{29}}{2}$
Work Step by Step
$\dfrac{7}{x^{2}-5x+6}=\dfrac{2x}{x-3}-\dfrac{x}{x-2}$
Factor the denominator of the fraction on the left side of the equation:
$\dfrac{7}{(x-2)(x-3)}=\dfrac{2x}{x-3}-\dfrac{x}{x-2}$
Multiply the whole equation by $(x-2)(x-3)$:
$(x-2)(x-3)\Big[\dfrac{7}{(x-2)(x-3)}=\dfrac{2x}{x-3}-\dfrac{x}{x-2}\Big]$
$7=2x(x-2)-x(x-3)$
$7=2x^{2}-4x-x^{2}+3x$
Take all terms to the right side of the equation and simplify it:
$0=2x^{2}-4x-x^{2}+3x-7$
$x^{2}-x-7=0$
Use the quadratic formula to solve this equation. The formula is $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Here, $a=1$, $b=-1$ and $c=-7$.
Substitute the known values into the formula and simplify:
$x=\dfrac{-(-1)\pm\sqrt{(-1)^{2}-4(1)(-7)}}{2(1)}=\dfrac{1\pm\sqrt{1+28}}{2}=...$
$...=\dfrac{1\pm\sqrt{29}}{2}=\dfrac{1}{2}\pm\dfrac{\sqrt{29}}{2}$
The original equation is not undefined for neither of the values of $x$ found. Our final answer is $x=\dfrac{1}{2}\pm\dfrac{\sqrt{29}}{2}$