X^2+y^2=40X+Y=4

Solution for X^2+y^2=40X+Y=4 equation:

X^2+X^2=40X+=4

We move all terms to the left:

X^2+X^2-(40X+)=0

We add all the numbers together, and all the variables

X^2+X^2-(+40X)=0

We add all the numbers together, and all the variables

2X^2-(+40X)=0

We get rid of parentheses

2X^2-40X=0

a = 2; b = -40; c = 0;
Δ = b2-4ac
Δ = -402-4·2·0
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-40}{2*2}=\frac{0}{4}
=0
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+40}{2*2}=\frac{80}{4}
=20
$