Y=3x^2+12x-96

Solution for Y=3x^2+12x-96 equation:

=3Y^2+12Y-96

We move all terms to the left:

-(3Y^2+12Y-96)=0

We get rid of parentheses

-3Y^2-12Y+96=0

a = -3; b = -12; c = +96;
Δ = b2-4ac
Δ = -122-4·(-3)·96
Δ = 1296
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1296}=36$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-36}{2*-3}=\frac{-24}{-6}
=+4
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+36}{2*-3}=\frac{48}{-6}
=-8
$