3y^2+4y+8=-6y

Solution for 3y^2+4y+8=-6y equation:

3y^2+4y+8=-6y

We move all terms to the left:

3y^2+4y+8-(-6y)=0

We get rid of parentheses

3y^2+4y+6y+8=0

We add all the numbers together, and all the variables

3y^2+10y+8=0

a = 3; b = 10; c = +8;
Δ = b2-4ac
Δ = 102-4·3·8
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2}{2*3}=\frac{-12}{6}
=-2
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2}{2*3}=\frac{-8}{6}
=-1+1/3
$