-3y^2+12y=-41

Solution for -3y^2+12y=-41 equation:

-3y^2+12y=-41

We move all terms to the left:

-3y^2+12y-(-41)=0

We add all the numbers together, and all the variables

-3y^2+12y+41=0

a = -3; b = 12; c = +41;
Δ = b2-4ac
Δ = 122-4·(-3)·41
Δ = 636
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:

$\sqrt{\Delta}=\sqrt{636}=\sqrt{4*159}=\sqrt{4}*\sqrt{159}=2\sqrt{159}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2\sqrt{159}}{2*-3}=\frac{-12-2\sqrt{159}}{-6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2\sqrt{159}}{2*-3}=\frac{-12+2\sqrt{159}}{-6}
$