4y-y^2=-41

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

4y-y^2=-41

We move all terms to the left:

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

We add all the numbers together, and all the variables

-1y^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}
$