0=3y^2+8y+4

Solution for 0=3y^2+8y+4 equation:

0=3y^2+8y+4

We move all terms to the left:

0-(3y^2+8y+4)=0

We add all the numbers together, and all the variables

-(3y^2+8y+4)=0

We get rid of parentheses

-3y^2-8y-4=0

a = -3; b = -8; c = -4;
Δ = b2-4ac
Δ = -82-4·(-3)·(-4)
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{-(-8)-4}{2*-3}=\frac{4}{-6}
=-2/3
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+4}{2*-3}=\frac{12}{-6}
=-2
$