(6y+4)(3y-5)=2y-3

Solution for (6y+4)(3y-5)=2y-3 equation:

(6y+4)(3y-5)=2y-3

We move all terms to the left:

(6y+4)(3y-5)-(2y-3)=0

We get rid of parentheses

(6y+4)(3y-5)-2y+3=0

We multiply parentheses ..

(+18y^2-30y+12y-20)-2y+3=0

We get rid of parentheses

18y^2-30y+12y-2y-20+3=0

We add all the numbers together, and all the variables

18y^2-20y-17=0

a = 18; b = -20; c = -17;
Δ = b2-4ac
Δ = -202-4·18·(-17)
Δ = 1624
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:

$\sqrt{\Delta}=\sqrt{1624}=\sqrt{4*406}=\sqrt{4}*\sqrt{406}=2\sqrt{406}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-2\sqrt{406}}{2*18}=\frac{20-2\sqrt{406}}{36}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+2\sqrt{406}}{2*18}=\frac{20+2\sqrt{406}}{36}
$