(4y+3)(2y-3)=45

Solution for (4y+3)(2y-3)=45 equation:

(4y+3)(2y-3)=45

We move all terms to the left:

(4y+3)(2y-3)-(45)=0

We multiply parentheses ..

(+8y^2-12y+6y-9)-45=0

We get rid of parentheses

8y^2-12y+6y-9-45=0

We add all the numbers together, and all the variables

8y^2-6y-54=0

a = 8; b = -6; c = -54;
Δ = b2-4ac
Δ = -62-4·8·(-54)
Δ = 1764
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1764}=42$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-42}{2*8}=\frac{-36}{16}
=-2+1/4
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+42}{2*8}=\frac{48}{16}
=3
$