32^2+x^2=38^2

Solution for 32^2+x^2=38^2 equation:

32^2+x^2=38^2

We move all terms to the left:

32^2+x^2-(38^2)=0

We add all the numbers together, and all the variables

x^2-420=0

a = 1; b = 0; c = -420;
Δ = b2-4ac
Δ = 02-4·1·(-420)
Δ = 1680
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:

$\sqrt{\Delta}=\sqrt{1680}=\sqrt{16*105}=\sqrt{16}*\sqrt{105}=4\sqrt{105}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{105}}{2*1}=\frac{0-4\sqrt{105}}{2}
=-\frac{4\sqrt{105}}{2}
=-2\sqrt{105}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{105}}{2*1}=\frac{0+4\sqrt{105}}{2}
=\frac{4\sqrt{105}}{2}
=2\sqrt{105}
$