6y(y-3)=-7y-4

Solution for 6y(y-3)=-7y-4 equation:

6y(y-3)=-7y-4

We move all terms to the left:

6y(y-3)-(-7y-4)=0

We multiply parentheses

6y^2-18y-(-7y-4)=0

We get rid of parentheses

6y^2-18y+7y+4=0

We add all the numbers together, and all the variables

6y^2-11y+4=0

a = 6; b = -11; c = +4;
Δ = b2-4ac
Δ = -112-4·6·4
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-5}{2*6}=\frac{6}{12}
=1/2
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+5}{2*6}=\frac{16}{12}
=1+1/3
$