Y=4x^2+8x-8

Solution for Y=4x^2+8x-8 equation:

=4Y^2+8Y-8

We move all terms to the left:

-(4Y^2+8Y-8)=0

We get rid of parentheses

-4Y^2-8Y+8=0

a = -4; b = -8; c = +8;
Δ = b2-4ac
Δ = -82-4·(-4)·8
Δ = 192
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:

$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-8\sqrt{3}}{2*-4}=\frac{8-8\sqrt{3}}{-8}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+8\sqrt{3}}{2*-4}=\frac{8+8\sqrt{3}}{-8}
$