Y=3x^2+24x+36

Solution for Y=3x^2+24x+36 equation:

=3Y^2+24Y+36

We move all terms to the left:

-(3Y^2+24Y+36)=0

We get rid of parentheses

-3Y^2-24Y-36=0

a = -3; b = -24; c = -36;
Δ = b2-4ac
Δ = -242-4·(-3)·(-36)
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-12}{2*-3}=\frac{12}{-6}
=-2
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+12}{2*-3}=\frac{36}{-6}
=-6
$