first of all you need push the restriction for what x value the denominator result 0 bc. than this fraction will be undefined
- so x^2 -25 = 0
(x-5)(x+5)=0
x-5=0 =>x=5
and
x+5=0 =>x=-5
so this mean that x not can being equal +/- 5
rewrite :
x^2 -25 = (x-5)(x+5)
2x 10 4
------------- = ------------------ - -----------
(x-5)(x+5) (x-5)(x+5) x +5
2x = 10 -4(x-5)
2x = 10 -4x +20
6x = 30
x = 30/6
x = 5 what is restriction for x
hope helped
