(x^2+10x+25)+(x^2-4x+4)=81

Solution for (x^2+10x+25)+(x^2-4x+4)=81 equation:

(x^2+10x+25)+(x^2-4x+4)=81

We move all terms to the left:

(x^2+10x+25)+(x^2-4x+4)-(81)=0

We get rid of parentheses

x^2+x^2+10x-4x+25+4-81=0

We add all the numbers together, and all the variables

2x^2+6x-52=0

a = 2; b = 6; c = -52;
Δ = b2-4ac
Δ = 62-4·2·(-52)
Δ = 452
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{452}=\sqrt{4*113}=\sqrt{4}*\sqrt{113}=2\sqrt{113}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{113}}{2*2}=\frac{-6-2\sqrt{113}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{113}}{2*2}=\frac{-6+2\sqrt{113}}{4}
$