19^2+b^2=21^2

Solution for 19^2+b^2=21^2 equation:

19^2+b^2=21^2

We move all terms to the left:

19^2+b^2-(21^2)=0

We add all the numbers together, and all the variables

b^2-80=0

a = 1; b = 0; c = -80;
Δ = b2-4ac
Δ = 02-4·1·(-80)
Δ = 320
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:

$\sqrt{\Delta}=\sqrt{320}=\sqrt{64*5}=\sqrt{64}*\sqrt{5}=8\sqrt{5}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{5}}{2*1}=\frac{0-8\sqrt{5}}{2}
=-\frac{8\sqrt{5}}{2}
=-4\sqrt{5}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{5}}{2*1}=\frac{0+8\sqrt{5}}{2}
=\frac{8\sqrt{5}}{2}
=4\sqrt{5}
$