Answer
Solution set = $\{-10\}$.
Work Step by Step
First, we exclude those values of x that yield a zero in any of the denominators.
$x\not\in\{-1,2 \}\qquad (*)$
Multiply the equation with the LCD=$(x+1)(x-2)$
$ 3x(x-2)+4(x+1)=3(x+1)(x-2)\qquad$ ... simplify (distribute)
$3x^{2}-6x+4x+4=3(x^{2}-x-2)$
$ 3x^{2}-2x+4=3x^{2}-3x-6\qquad$ ... add $-4-3x^{2}+3x$
$-2x+3x=-6-4$
$ x=-10 \qquad$ ...Checking (*), this is a valid solution
Solution set = $\{-10\}$.
