Y=18x^2+96x-12

Solution for Y=18x^2+96x-12 equation:

=18Y^2+96Y-12

We move all terms to the left:

-(18Y^2+96Y-12)=0

We get rid of parentheses

-18Y^2-96Y+12=0

a = -18; b = -96; c = +12;
Δ = b2-4ac
Δ = -962-4·(-18)·12
Δ = 10080
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:

$\sqrt{\Delta}=\sqrt{10080}=\sqrt{144*70}=\sqrt{144}*\sqrt{70}=12\sqrt{70}$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-96)-12\sqrt{70}}{2*-18}=\frac{96-12\sqrt{70}}{-36}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-96)+12\sqrt{70}}{2*-18}=\frac{96+12\sqrt{70}}{-36}
$