Answer
$\left\{ \dfrac{1-\sqrt{29}}{2},\dfrac{1+\sqrt{29}}{2} \right\}$
Work Step by Step
Factoring the given expression and then multiplying both sides of the given equation, $
\dfrac{7}{x^2-5x+6}=\dfrac{2x}{x-3}-\dfrac{x}{x-2}
,$ by the $LCD=
(x-3)(x-2)
$, then,
\begin{array}{l}\require{cancel}
\dfrac{7}{(x-3)(x-2)}=\dfrac{2x}{x-3}-\dfrac{x}{x-2}
\\\\
(x-3)(x-2)\left( \dfrac{7}{(x-3)(x-2)} \right)=\left( \dfrac{2x}{x-3}-\dfrac{x}{x-2} \right)(x-3)(x-2)
\\\\
1(7)=2x(x-2)-x(x-3)
\\\\
7=2x^2-4x-x^2+3x
\\\\
(-2x^2+x^2)+(4x-3x)+7=0
\\\\
-x^2+x+7=0
\\\\
x^2-x-7=0
.\end{array}
Using $\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ (or the Quadratic Formula), the solutions of the quadratic equation, $
x^2-x-7=0
,$ are
\begin{array}{l}\require{cancel}
\dfrac{-(-1)\pm\sqrt{(-1)^2-4(1)(-7)}}{2(1)}
\\\\=
\dfrac{1\pm\sqrt{1+28}}{2}
\\\\=
\dfrac{1\pm\sqrt{29}}{2}
.\end{array}
Upon checking, both solutions satisfy the original equation. Hence, the solutions are $
\left\{ \dfrac{1-\sqrt{29}}{2},\dfrac{1+\sqrt{29}}{2} \right\}
.$
