198-40y+2y^2=0

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

198-40y+2y^2=0

a = 2; b = -40; c = +198;
Δ = b2-4ac
Δ = -402-4·2·198
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-4}{2*2}=\frac{36}{4}
=9
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+4}{2*2}=\frac{44}{4}
=11
$