Y^3+2y^2+16y+32=0

Solution for Y^3+2y^2+16y+32=0 equation:

^3+2Y^2+16Y+32=0

We add all the numbers together, and all the variables

2Y^2+16Y=0

a = 2; b = 16; c = 0;
Δ = b2-4ac
Δ = 162-4·2·0
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{-(16)-16}{2*2}=\frac{-32}{4}
=-8
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+16}{2*2}=\frac{0}{4}
=0
$