20^2+x^2=40^2

Solution for 20^2+x^2=40^2 equation:

20^2+x^2=40^2

We move all terms to the left:

20^2+x^2-(40^2)=0

We add all the numbers together, and all the variables

x^2-1200=0

a = 1; b = 0; c = -1200;
Δ = b2-4ac
Δ = 02-4·1·(-1200)
Δ = 4800
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:

$\sqrt{\Delta}=\sqrt{4800}=\sqrt{1600*3}=\sqrt{1600}*\sqrt{3}=40\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-40\sqrt{3}}{2*1}=\frac{0-40\sqrt{3}}{2}
=-\frac{40\sqrt{3}}{2}
=-20\sqrt{3}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+40\sqrt{3}}{2*1}=\frac{0+40\sqrt{3}}{2}
=\frac{40\sqrt{3}}{2}
=20\sqrt{3}
$