y^2-4y=41

Solution for y^2-4y=41 equation:

y^2-4y=41

We move all terms to the left:

y^2-4y-(41)=0

a = 1; b = -4; c = -41;
Δ = b2-4ac
Δ = -42-4·1·(-41)
Δ = 180
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{180}=\sqrt{36*5}=\sqrt{36}*\sqrt{5}=6\sqrt{5}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-6\sqrt{5}}{2*1}=\frac{4-6\sqrt{5}}{2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+6\sqrt{5}}{2*1}=\frac{4+6\sqrt{5}}{2}
$