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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We multiply parentheses ..

(+8y^2+12y-2y-3)-18y+4=0

We get rid of parentheses

8y^2+12y-2y-18y-3+4=0

We add all the numbers together, and all the variables

8y^2-8y+1=0

a = 8; b = -8; c = +1;
Δ = b2-4ac
Δ = -82-4·8·1
Δ = 32
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:

$\sqrt{\Delta}=\sqrt{32}=\sqrt{16*2}=\sqrt{16}*\sqrt{2}=4\sqrt{2}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-4\sqrt{2}}{2*8}=\frac{8-4\sqrt{2}}{16}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+4\sqrt{2}}{2*8}=\frac{8+4\sqrt{2}}{16}
$