Y=20x^2+36x-35

Solution for Y=20x^2+36x-35 equation:

=20Y^2+36Y-35

We move all terms to the left:

-(20Y^2+36Y-35)=0

We get rid of parentheses

-20Y^2-36Y+35=0

a = -20; b = -36; c = +35;
Δ = b2-4ac
Δ = -362-4·(-20)·35
Δ = 4096
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4096}=64$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-36)-64}{2*-20}=\frac{-28}{-40}
=7/10
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-36)+64}{2*-20}=\frac{100}{-40}
=-2+1/2
$