X2+y2=212

Solution for X2+y2=212 equation:

X2+X2=212

We move all terms to the left:

X2+X2-(212)=0

We add all the numbers together, and all the variables

2X^2-212=0

a = 2; b = 0; c = -212;
Δ = b2-4ac
Δ = 02-4·2·(-212)
Δ = 1696
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:

$\sqrt{\Delta}=\sqrt{1696}=\sqrt{16*106}=\sqrt{16}*\sqrt{106}=4\sqrt{106}$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{106}}{2*2}=\frac{0-4\sqrt{106}}{4}
=-\frac{4\sqrt{106}}{4}
=-\sqrt{106}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{106}}{2*2}=\frac{0+4\sqrt{106}}{4}
=\frac{4\sqrt{106}}{4}
=\sqrt{106}
$