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

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

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

We move all terms to the left:

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

We multiply parentheses ..

4(+6y^2+4y-3y-2)-1=0

We multiply parentheses

24y^2+16y-12y-8-1=0

We add all the numbers together, and all the variables

24y^2+4y-9=0

a = 24; b = 4; c = -9;
Δ = b2-4ac
Δ = 42-4·24·(-9)
Δ = 880
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:

$\sqrt{\Delta}=\sqrt{880}=\sqrt{16*55}=\sqrt{16}*\sqrt{55}=4\sqrt{55}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{55}}{2*24}=\frac{-4-4\sqrt{55}}{48}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{55}}{2*24}=\frac{-4+4\sqrt{55}}{48}
$