100-x^2=40-4x+40+4x

Solution for 100-x^2=40-4x+40+4x equation:

100-x^2=40-4x+40+4x

We move all terms to the left:

100-x^2-(40-4x+40+4x)=0

We add all the numbers together, and all the variables

-x^2+100-80=0

We add all the numbers together, and all the variables

-1x^2+20=0

a = -1; b = 0; c = +20;
Δ = b2-4ac
Δ = 02-4·(-1)·20
Δ = 80
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:

$\sqrt{\Delta}=\sqrt{80}=\sqrt{16*5}=\sqrt{16}*\sqrt{5}=4\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{5}}{2*-1}=\frac{0-4\sqrt{5}}{-2}
=-\frac{4\sqrt{5}}{-2}
=-\frac{2\sqrt{5}}{-1}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{5}}{2*-1}=\frac{0+4\sqrt{5}}{-2}
=\frac{4\sqrt{5}}{-2}
=\frac{2\sqrt{5}}{-1}
$