19^2+19^2=x^2

Solution for 19^2+19^2=x^2 equation:

19^2+19^2=x^2

We move all terms to the left:

19^2+19^2-(x^2)=0

We add all the numbers together, and all the variables

-1x^2+722=0

a = -1; b = 0; c = +722;
Δ = b2-4ac
Δ = 02-4·(-1)·722
Δ = 2888
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:

$\sqrt{\Delta}=\sqrt{2888}=\sqrt{1444*2}=\sqrt{1444}*\sqrt{2}=38\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-38\sqrt{2}}{2*-1}=\frac{0-38\sqrt{2}}{-2}
=-\frac{38\sqrt{2}}{-2}
=-\frac{19\sqrt{2}}{-1}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+38\sqrt{2}}{2*-1}=\frac{0+38\sqrt{2}}{-2}
=\frac{38\sqrt{2}}{-2}
=\frac{19\sqrt{2}}{-1}
$