(x^2)+(x^2)=128

Solution for (x^2)+(x^2)=128 equation:

(x^2)+(x^2)=128

We move all terms to the left:

(x^2)+(x^2)-(128)=0

We add all the numbers together, and all the variables

2x^2-128=0

a = 2; b = 0; c = -128;
Δ = b2-4ac
Δ = 02-4·2·(-128)
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-32}{2*2}=\frac{-32}{4}
=-8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+32}{2*2}=\frac{32}{4}
=8
$