13^2+x^2=19^2

Solution for 13^2+x^2=19^2 equation:

13^2+x^2=19^2

We move all terms to the left:

13^2+x^2-(19^2)=0

We add all the numbers together, and all the variables

x^2-192=0

a = 1; b = 0; c = -192;
Δ = b2-4ac
Δ = 02-4·1·(-192)
Δ = 768
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:

$\sqrt{\Delta}=\sqrt{768}=\sqrt{256*3}=\sqrt{256}*\sqrt{3}=16\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{3}}{2*1}=\frac{0-16\sqrt{3}}{2}
=-\frac{16\sqrt{3}}{2}
=-8\sqrt{3}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{3}}{2*1}=\frac{0+16\sqrt{3}}{2}
=\frac{16\sqrt{3}}{2}
=8\sqrt{3}
$