Y=2x^2+20x+40

Solution for Y=2x^2+20x+40 equation:

=2Y^2+20Y+40

We move all terms to the left:

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

We get rid of parentheses

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

a = -2; b = -20; c = -40;
Δ = b2-4ac
Δ = -202-4·(-2)·(-40)
Δ = 80
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:

$\sqrt{\Delta}=\sqrt{80}=\sqrt{16*5}=\sqrt{16}*\sqrt{5}=4\sqrt{5}$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{5}}{2*-2}=\frac{20-4\sqrt{5}}{-4}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{5}}{2*-2}=\frac{20+4\sqrt{5}}{-4}
$