x^2+(x^2+4x+4)=100

Solution for x^2+(x^2+4x+4)=100 equation:

x^2+(x^2+4x+4)=100

We move all terms to the left:

x^2+(x^2+4x+4)-(100)=0

We get rid of parentheses

x^2+x^2+4x+4-100=0

We add all the numbers together, and all the variables

2x^2+4x-96=0

a = 2; b = 4; c = -96;
Δ = b2-4ac
Δ = 42-4·2·(-96)
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-28}{2*2}=\frac{-32}{4}
=-8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+28}{2*2}=\frac{24}{4}
=6
$