x^2+21^2=37^2

Solution for x^2+21^2=37^2 equation:

x^2+21^2=37^2

We move all terms to the left:

x^2+21^2-(37^2)=0

We add all the numbers together, and all the variables

x^2-928=0

a = 1; b = 0; c = -928;
Δ = b2-4ac
Δ = 02-4·1·(-928)
Δ = 3712
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:

$\sqrt{\Delta}=\sqrt{3712}=\sqrt{64*58}=\sqrt{64}*\sqrt{58}=8\sqrt{58}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{58}}{2*1}=\frac{0-8\sqrt{58}}{2}
=-\frac{8\sqrt{58}}{2}
=-4\sqrt{58}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{58}}{2*1}=\frac{0+8\sqrt{58}}{2}
=\frac{8\sqrt{58}}{2}
=4\sqrt{58}
$