Answer
Any value of x that cause a zero in any of the denominators must be excluded, because the equation is not defined for these values.
The LHS has denominators (x+5) and (x-2), so we must exclude $ x=-5$ and $ x=2$.
The RHS has $ x^{2}+3x-6$ as a denominator.
Find zeros with the quadratic formula:
$ x=\displaystyle \frac{-3\pm\sqrt{9-4(1)(-6)}}{2(1)}= \displaystyle \frac{-3\pm\sqrt{33}}{2}$
So, the numbers $ x= \displaystyle \frac{-3-\sqrt{33}}{2}$ and $ x= \displaystyle \frac{-3+\sqrt{33}}{2}$
are also excluded.
Work Step by Step
Given above.
