(6y+4)(3y-5)=4

Solution for (6y+4)(3y-5)=4 equation:

(6y+4)(3y-5)=4

We move all terms to the left:

(6y+4)(3y-5)-(4)=0

We multiply parentheses ..

(+18y^2-30y+12y-20)-4=0

We get rid of parentheses

18y^2-30y+12y-20-4=0

We add all the numbers together, and all the variables

18y^2-18y-24=0

a = 18; b = -18; c = -24;
Δ = b2-4ac
Δ = -182-4·18·(-24)
Δ = 2052
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{2052}=\sqrt{36*57}=\sqrt{36}*\sqrt{57}=6\sqrt{57}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-6\sqrt{57}}{2*18}=\frac{18-6\sqrt{57}}{36}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+6\sqrt{57}}{2*18}=\frac{18+6\sqrt{57}}{36}
$