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

Solution for 2(4y-3)(6y+1)=0 equation:

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

We multiply parentheses ..

2(+24y^2+4y-18y-3)=0

We multiply parentheses

48y^2+8y-36y-6=0

We add all the numbers together, and all the variables

48y^2-28y-6=0

a = 48; b = -28; c = -6;
Δ = b2-4ac
Δ = -282-4·48·(-6)
Δ = 1936
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1936}=44$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-44}{2*48}=\frac{-16}{96}
=-1/6
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+44}{2*48}=\frac{72}{96}
=3/4
$