Answer
Solution set = $\{7\}$.
Work Step by Step
First, we exclude those values of x that yield a zero in any of the denominators.
$x\not\in\{-5,+5 \}\qquad (*)$
The first denominator is a difference of squares, $(x+5)(x-5)$
Multiply the equation with the LCD=$(x+5)(x-5)$
$ 32=4(x-5)+2(x+5)\qquad$ ... simplify (distribute)
$32=4x-20+2x+10$
$ 32=6x-10\qquad\qquad $ ... add $10$
$42=6x$
$ x=7 \qquad$ ...Checking (*), this is a valid solution
Solution set = $\{7\}$.
