(2y+3)(2y-3)=4y+9

Solution for (2y+3)(2y-3)=4y+9 equation:

(2y+3)(2y-3)=4y+9

We move all terms to the left:

(2y+3)(2y-3)-(4y+9)=0

We use the square of the difference formula

4y^2-(4y+9)-9=0

We get rid of parentheses

4y^2-4y-9-9=0

We add all the numbers together, and all the variables

4y^2-4y-18=0

a = 4; b = -4; c = -18;
Δ = b2-4ac
Δ = -42-4·4·(-18)
Δ = 304
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:

$\sqrt{\Delta}=\sqrt{304}=\sqrt{16*19}=\sqrt{16}*\sqrt{19}=4\sqrt{19}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{19}}{2*4}=\frac{4-4\sqrt{19}}{8}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{19}}{2*4}=\frac{4+4\sqrt{19}}{8}
$