-3y^2+48y-2=0

Solution for -3y^2+48y-2=0 equation:

-3y^2+48y-2=0

a = -3; b = 48; c = -2;
Δ = b2-4ac
Δ = 482-4·(-3)·(-2)
Δ = 2280
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:

$\sqrt{\Delta}=\sqrt{2280}=\sqrt{4*570}=\sqrt{4}*\sqrt{570}=2\sqrt{570}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-2\sqrt{570}}{2*-3}=\frac{-48-2\sqrt{570}}{-6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+2\sqrt{570}}{2*-3}=\frac{-48+2\sqrt{570}}{-6}
$