3y^2+y+2y=0

Solution for 3y^2+y+2y=0 equation:

3y^2+y+2y=0

We add all the numbers together, and all the variables

3y^2+3y=0

a = 3; b = 3; c = 0;
Δ = b2-4ac
Δ = 32-4·3·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3}{2*3}=\frac{-6}{6}
=-1
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3}{2*3}=\frac{0}{6}
=0
$