3y-3y-3+6-4y2=0

Solution for 3y-3y-3+6-4y2=0 equation:

3y-3y-3+6-4y^2=0

We add all the numbers together, and all the variables

-4y^2+3=0

a = -4; b = 0; c = +3;
Δ = b2-4ac
Δ = 02-4·(-4)·3
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{-(0)-4\sqrt{3}}{2*-4}=\frac{0-4\sqrt{3}}{-8}
=-\frac{4\sqrt{3}}{-8}
=-\frac{\sqrt{3}}{-2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{3}}{2*-4}=\frac{0+4\sqrt{3}}{-8}
=\frac{4\sqrt{3}}{-8}
=\frac{\sqrt{3}}{-2}
$