45=(2x-3)(2x+1)

Solution for 45=(2x-3)(2x+1) equation:

45=(2x-3)(2x+1)

We move all terms to the left:

45-((2x-3)(2x+1))=0

We multiply parentheses ..

-((+4x^2+2x-6x-3))+45=0


We calculate terms in parentheses: -((+4x^2+2x-6x-3)), so:

(+4x^2+2x-6x-3)

We get rid of parentheses

4x^2+2x-6x-3

We add all the numbers together, and all the variables

4x^2-4x-3

Back to the equation:

-(4x^2-4x-3)


We get rid of parentheses

-4x^2+4x+3+45=0

We add all the numbers together, and all the variables

-4x^2+4x+48=0

a = -4; b = 4; c = +48;
Δ = b2-4ac
Δ = 42-4·(-4)·48
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-28}{2*-4}=\frac{-32}{-8}
=+4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+28}{2*-4}=\frac{24}{-8}
=-3
$