X^2+10x-7 / x^2+10x+25=0

how do i solve for x?

Factorize both the numerator and denominator then equate to zero ... you'll get X

You just have to solve the next equation:

x²+10x-7=0

Because, if the numerator is equal to "0", then the next expression

(x²+10x-7) / (x²+10x+25) = 0

Therefore:

x²+10x-7=0

x=[-10⁺₋√ (100+28) ]/2 = (-10⁺₋√128) / 2 = (-10⁺₋8√2) / 2=-5⁺₋4√2

We have two possible solutions:

x₁=-5-4√2

x₂=-5+4√2

Now, we have to check these possible solutions.

Remember the denominator cannot be equal to "0".

if x=-5-4√2

[ (-5-4√2) ²+10 (-5-4√2) - 7] / [ (-5-4√2) ²+10 (-5-4√2) + 25]=0/32=0

Therefore: x=-5-4√2 is a solution.

if x=-5+4√2

[ (-5+4√2) ²+10 (-5+4√2) - 7] / [ (-5+4√2) ²+10 (-5+4√2) + 25]=0/32=0

Therefore: x=-5+4√√2 is a solution.

Answer: we have two solutions for x:

x₁=-5-4√2

x₂=-5+4√2