4y(3y+8)=3

Solution for 4y(3y+8)=3 equation:

4y(3y+8)=3

We move all terms to the left:

4y(3y+8)-(3)=0

We multiply parentheses

12y^2+32y-3=0

a = 12; b = 32; c = -3;
Δ = b2-4ac
Δ = 322-4·12·(-3)
Δ = 1168
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:

$\sqrt{\Delta}=\sqrt{1168}=\sqrt{16*73}=\sqrt{16}*\sqrt{73}=4\sqrt{73}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-4\sqrt{73}}{2*12}=\frac{-32-4\sqrt{73}}{24}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+4\sqrt{73}}{2*12}=\frac{-32+4\sqrt{73}}{24}
$