2y^2+40y-198=0

Solution for 2y^2+40y-198=0 equation:

2y^2+40y-198=0

a = 2; b = 40; c = -198;
Δ = b2-4ac
Δ = 402-4·2·(-198)
Δ = 3184
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:

$\sqrt{\Delta}=\sqrt{3184}=\sqrt{16*199}=\sqrt{16}*\sqrt{199}=4\sqrt{199}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-4\sqrt{199}}{2*2}=\frac{-40-4\sqrt{199}}{4}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+4\sqrt{199}}{2*2}=\frac{-40+4\sqrt{199}}{4}
$