Y=20x2+13x+-40

Solution for Y=20x2+13x+-40 equation:

=20Y^2+13Y+-40

We move all terms to the left:

-(20Y^2+13Y+-40)=0

We use the square of the difference formula

-(20Y^2+13Y-40)=0

We get rid of parentheses

-20Y^2-13Y+40=0

a = -20; b = -13; c = +40;
Δ = b2-4ac
Δ = -132-4·(-20)·40
Δ = 3369
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-\sqrt{3369}}{2*-20}=\frac{13-\sqrt{3369}}{-40}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+\sqrt{3369}}{2*-20}=\frac{13+\sqrt{3369}}{-40}
$