0=(2a-3)(2a+1)

Solution for 0=(2a-3)(2a+1) equation:

0=(2a-3)(2a+1)

We move all terms to the left:

0-((2a-3)(2a+1))=0

We add all the numbers together, and all the variables

-((2a-3)(2a+1))=0

We multiply parentheses ..

-((+4a^2+2a-6a-3))=0


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

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

We get rid of parentheses

4a^2+2a-6a-3

We add all the numbers together, and all the variables

4a^2-4a-3

Back to the equation:

-(4a^2-4a-3)


We get rid of parentheses

-4a^2+4a+3=0

a = -4; b = 4; c = +3;
Δ = b2-4ac
Δ = 42-4·(-4)·3
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-8}{2*-4}=\frac{-12}{-8}
=1+1/2
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+8}{2*-4}=\frac{4}{-8}
=-1/2
$