Y=4x^2+36x+80

Solution for Y=4x^2+36x+80 equation:

=4Y^2+36Y+80

We move all terms to the left:

-(4Y^2+36Y+80)=0

We get rid of parentheses

-4Y^2-36Y-80=0

a = -4; b = -36; c = -80;
Δ = b2-4ac
Δ = -362-4·(-4)·(-80)
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-36)-4}{2*-4}=\frac{32}{-8}
=-4
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-36)+4}{2*-4}=\frac{40}{-8}
=-5
$