Answer
$\{ -\displaystyle \frac{5}{2},1\}$
Work Step by Step
We exclude $x=-4$ and $x=2$ from the solutions,
as the equation would not be defined.
Multiply with the LCM, $(x+4)(x-2)$
$5(x-2)=4(x+4)(x-2) +3(x+4)$
Distribute and write in standard form.
$5x-10=4(x^{2}+2x-8)+3x+12$
$5x-10=4x^{2}+8x-32+3x+12$
$5x-10=4x^{2}+11x-20$
$0=4x^{2}+6x-10$
...divide with 2...
$2x^{2}+3x-5=0$
We search for two factors of $(ac=-10)$, whose sum is $b=+3.$
... we find $-2$ and $+5.$ Rewrite the LHS:
$2x^{2}-2x+5x-5=0$
... factor in pairs
$2x(x-1)+5(x-1)=0$
$(2x+5)(x-1)=0$
Either
$2x+5=0 \Rightarrow x=-\displaystyle \frac{5}{2}$
or $x=1$
(the restrictrions are not violated for either x)
Solution set = $\{ -\displaystyle \frac{5}{2},1\}$
