(4y+3)(2y-3)=36

Solution for (4y+3)(2y-3)=36 equation:

(4y+3)(2y-3)=36

We move all terms to the left:

(4y+3)(2y-3)-(36)=0

We multiply parentheses ..

(+8y^2-12y+6y-9)-36=0

We get rid of parentheses

8y^2-12y+6y-9-36=0

We add all the numbers together, and all the variables

8y^2-6y-45=0

a = 8; b = -6; c = -45;
Δ = b2-4ac
Δ = -62-4·8·(-45)
Δ = 1476
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:

$\sqrt{\Delta}=\sqrt{1476}=\sqrt{36*41}=\sqrt{36}*\sqrt{41}=6\sqrt{41}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-6\sqrt{41}}{2*8}=\frac{6-6\sqrt{41}}{16}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+6\sqrt{41}}{2*8}=\frac{6+6\sqrt{41}}{16}
$