58^2+40^2=b^2

Solution for 58^2+40^2=b^2 equation:

58^2+40^2=b^2

We move all terms to the left:

58^2+40^2-(b^2)=0

We add all the numbers together, and all the variables

-1b^2+4964=0

a = -1; b = 0; c = +4964;
Δ = b2-4ac
Δ = 02-4·(-1)·4964
Δ = 19856
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{19856}=\sqrt{16*1241}=\sqrt{16}*\sqrt{1241}=4\sqrt{1241}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{1241}}{2*-1}=\frac{0-4\sqrt{1241}}{-2}
=-\frac{4\sqrt{1241}}{-2}
=-\frac{2\sqrt{1241}}{-1}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{1241}}{2*-1}=\frac{0+4\sqrt{1241}}{-2}
=\frac{4\sqrt{1241}}{-2}
=\frac{2\sqrt{1241}}{-1}
$