x^2+x^2=161

Solution for x^2+x^2=161 equation:

x^2+x^2=161

We move all terms to the left:

x^2+x^2-(161)=0

We add all the numbers together, and all the variables

2x^2-161=0

a = 2; b = 0; c = -161;
Δ = b2-4ac
Δ = 02-4·2·(-161)
Δ = 1288
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:

$\sqrt{\Delta}=\sqrt{1288}=\sqrt{4*322}=\sqrt{4}*\sqrt{322}=2\sqrt{322}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{322}}{2*2}=\frac{0-2\sqrt{322}}{4}
=-\frac{2\sqrt{322}}{4}
=-\frac{\sqrt{322}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{322}}{2*2}=\frac{0+2\sqrt{322}}{4}
=\frac{2\sqrt{322}}{4}
=\frac{\sqrt{322}}{2}
$