3y^2=-7y-4

Solution for 3y^2=-7y-4 equation:

3y^2=-7y-4

We move all terms to the left:

3y^2-(-7y-4)=0

We get rid of parentheses

3y^2+7y+4=0

a = 3; b = 7; c = +4;
Δ = b2-4ac
Δ = 72-4·3·4
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-1}{2*3}=\frac{-8}{6}
=-1+1/3
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+1}{2*3}=\frac{-6}{6}
=-1
$