Y=4x^2+36x-16

Solution for Y=4x^2+36x-16 equation:

=4Y^2+36Y-16

We move all terms to the left:

-(4Y^2+36Y-16)=0

We get rid of parentheses

-4Y^2-36Y+16=0

a = -4; b = -36; c = +16;
Δ = b2-4ac
Δ = -362-4·(-4)·16
Δ = 1552
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:

$\sqrt{\Delta}=\sqrt{1552}=\sqrt{16*97}=\sqrt{16}*\sqrt{97}=4\sqrt{97}$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-36)-4\sqrt{97}}{2*-4}=\frac{36-4\sqrt{97}}{-8}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-36)+4\sqrt{97}}{2*-4}=\frac{36+4\sqrt{97}}{-8}
$