(3y-1)(2y+4)=18

Solution for (3y-1)(2y+4)=18 equation:

(3y-1)(2y+4)=18

We move all terms to the left:

(3y-1)(2y+4)-(18)=0

We multiply parentheses ..

(+6y^2+12y-2y-4)-18=0

We get rid of parentheses

6y^2+12y-2y-4-18=0

We add all the numbers together, and all the variables

6y^2+10y-22=0

a = 6; b = 10; c = -22;
Δ = b2-4ac
Δ = 102-4·6·(-22)
Δ = 628
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:

$\sqrt{\Delta}=\sqrt{628}=\sqrt{4*157}=\sqrt{4}*\sqrt{157}=2\sqrt{157}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{157}}{2*6}=\frac{-10-2\sqrt{157}}{12}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{157}}{2*6}=\frac{-10+2\sqrt{157}}{12}
$