2+24y+40y^2=0

Solution for 2+24y+40y^2=0 equation:

2+24y+40y^2=0

a = 40; b = 24; c = +2;
Δ = b2-4ac
Δ = 242-4·40·2
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-16}{2*40}=\frac{-40}{80}
=-1/2
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+16}{2*40}=\frac{-8}{80}
=-1/10
$