X2+y2=41y=5

Solution for X2+y2=41y=5 equation:

X2+X2=41X=5

We move all terms to the left:

X2+X2-(41X)=0

We add all the numbers together, and all the variables

2X^2-41X=0

a = 2; b = -41; c = 0;
Δ = b2-4ac
Δ = -412-4·2·0
Δ = 1681
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1681}=41$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-41)-41}{2*2}=\frac{0}{4}
=0
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-41)+41}{2*2}=\frac{82}{4}
=20+1/2
$