Y=-2x^2+40x+8

Solution for Y=-2x^2+40x+8 equation:

=-2Y^2+40Y+8

We move all terms to the left:

-(-2Y^2+40Y+8)=0

We get rid of parentheses

2Y^2-40Y-8=0

a = 2; b = -40; c = -8;
Δ = b2-4ac
Δ = -402-4·2·(-8)
Δ = 1664
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:

$\sqrt{\Delta}=\sqrt{1664}=\sqrt{64*26}=\sqrt{64}*\sqrt{26}=8\sqrt{26}$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-8\sqrt{26}}{2*2}=\frac{40-8\sqrt{26}}{4}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+8\sqrt{26}}{2*2}=\frac{40+8\sqrt{26}}{4}
$