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

Solution for -2y^2+12y-12=0 equation:

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

a = -2; b = 12; c = -12;
Δ = b2-4ac
Δ = 122-4·(-2)·(-12)
Δ = 48
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:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4\sqrt{3}}{2*-2}=\frac{-12-4\sqrt{3}}{-4}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{3}}{2*-2}=\frac{-12+4\sqrt{3}}{-4}
$