(x^2 + 4x - 1)/(3 - 2x) = x
Solve for x: $\displaystyle \dfrac{x^{2}+4x-1}{-2x+3}={x}$.
Multiply both sides of the equation by $\displaystyle -2x+3$ to get:
$\displaystyle x^{2}+4x-1={-2x^{2}+3x}$.
Put all terms on one side of the equation, now we have:
$\displaystyle 3x^2+x-1=0$.
Use the quadratic formula:
$\displaystyle x={ \dfrac{-b \pm \sqrt{b^{2}-4ac}}{2a}} = { \dfrac{-(1) \pm \sqrt{(1)^{2}-4(3)(-1)}}{2(3)}} = {\dfrac{-1 \pm \sqrt{13}}{6}}$.
So $\displaystyle x = {\dfrac{-1 \pm \sqrt{13}}{6}}$.