(2y+3)4y=3

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

(2y+3)4y=3

We move all terms to the left:

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

We multiply parentheses

8y^2+12y-3=0

a = 8; b = 12; c = -3;
Δ = b2-4ac
Δ = 122-4·8·(-3)
Δ = 240
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:

$\sqrt{\Delta}=\sqrt{240}=\sqrt{16*15}=\sqrt{16}*\sqrt{15}=4\sqrt{15}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4\sqrt{15}}{2*8}=\frac{-12-4\sqrt{15}}{16}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{15}}{2*8}=\frac{-12+4\sqrt{15}}{16}
$