20^2+x^2=29^2

Solution for 20^2+x^2=29^2 equation:

20^2+x^2=29^2

We move all terms to the left:

20^2+x^2-(29^2)=0

We add all the numbers together, and all the variables

x^2-441=0

a = 1; b = 0; c = -441;
Δ = b2-4ac
Δ = 02-4·1·(-441)
Δ = 1764
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1764}=42$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-42}{2*1}=\frac{-42}{2}
=-21
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+42}{2*1}=\frac{42}{2}
=21
$