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

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

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

We move all terms to the left:

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

determiningTheFunctionDomain
-x^2+40^2+20^2=0

We add all the numbers together, and all the variables

-1x^2+2000=0

a = -1; b = 0; c = +2000;
Δ = b2-4ac
Δ = 02-4·(-1)·2000
Δ = 8000
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:

$\sqrt{\Delta}=\sqrt{8000}=\sqrt{1600*5}=\sqrt{1600}*\sqrt{5}=40\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-40\sqrt{5}}{2*-1}=\frac{0-40\sqrt{5}}{-2}
=-\frac{40\sqrt{5}}{-2}
=-\frac{20\sqrt{5}}{-1}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+40\sqrt{5}}{2*-1}=\frac{0+40\sqrt{5}}{-2}
=\frac{40\sqrt{5}}{-2}
=\frac{20\sqrt{5}}{-1}
$