Y=3x^2-3x-36

Solution for Y=3x^2-3x-36 equation:

=3Y^2-3Y-36

We move all terms to the left:

-(3Y^2-3Y-36)=0

We get rid of parentheses

-3Y^2+3Y+36=0

a = -3; b = 3; c = +36;
Δ = b2-4ac
Δ = 32-4·(-3)·36
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-21}{2*-3}=\frac{-24}{-6}
=+4
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+21}{2*-3}=\frac{18}{-6}
=-3
$