Y=3x^2-3x-20

Solution for Y=3x^2-3x-20 equation:

=3Y^2-3Y-20

We move all terms to the left:

-(3Y^2-3Y-20)=0

We get rid of parentheses

-3Y^2+3Y+20=0

a = -3; b = 3; c = +20;
Δ = b2-4ac
Δ = 32-4·(-3)·20
Δ = 249
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{249}}{2*-3}=\frac{-3-\sqrt{249}}{-6}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{249}}{2*-3}=\frac{-3+\sqrt{249}}{-6}
$