20^2+b^2=24^2

Solution for 20^2+b^2=24^2 equation:

20^2+b^2=24^2

We move all terms to the left:

20^2+b^2-(24^2)=0

We add all the numbers together, and all the variables

b^2-176=0

a = 1; b = 0; c = -176;
Δ = b2-4ac
Δ = 02-4·1·(-176)
Δ = 704
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:

$\sqrt{\Delta}=\sqrt{704}=\sqrt{64*11}=\sqrt{64}*\sqrt{11}=8\sqrt{11}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{11}}{2*1}=\frac{0-8\sqrt{11}}{2}
=-\frac{8\sqrt{11}}{2}
=-4\sqrt{11}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{11}}{2*1}=\frac{0+8\sqrt{11}}{2}
=\frac{8\sqrt{11}}{2}
=4\sqrt{11}
$