360=40y+2y^2

Solution for 360=40y+2y^2 equation:

360=40y+2y^2

We move all terms to the left:

360-(40y+2y^2)=0

We get rid of parentheses

-2y^2-40y+360=0

a = -2; b = -40; c = +360;
Δ = b2-4ac
Δ = -402-4·(-2)·360
Δ = 4480
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:

$\sqrt{\Delta}=\sqrt{4480}=\sqrt{64*70}=\sqrt{64}*\sqrt{70}=8\sqrt{70}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-8\sqrt{70}}{2*-2}=\frac{40-8\sqrt{70}}{-4}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+8\sqrt{70}}{2*-2}=\frac{40+8\sqrt{70}}{-4}
$