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

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

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

We move all terms to the left:

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

We multiply parentheses

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

8y^2+8y-3=0

a = 8; b = 8; c = -3;
Δ = b2-4ac
Δ = 82-4·8·(-3)
Δ = 160
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:

$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4\sqrt{10}}{2*8}=\frac{-8-4\sqrt{10}}{16}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4\sqrt{10}}{2*8}=\frac{-8+4\sqrt{10}}{16}
$