0=-4(2x+1)(2x-3)

Solution for 0=-4(2x+1)(2x-3) equation:

0=-4(2x+1)(2x-3)

We move all terms to the left:

0-(-4(2x+1)(2x-3))=0

We add all the numbers together, and all the variables

-(-4(2x+1)(2x-3))=0

We multiply parentheses ..

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


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

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

We multiply parentheses

-16x^2+24x-8x+12

We add all the numbers together, and all the variables

-16x^2+16x+12

Back to the equation:

-(-16x^2+16x+12)


We get rid of parentheses

16x^2-16x-12=0

a = 16; b = -16; c = -12;
Δ = b2-4ac
Δ = -162-4·16·(-12)
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-32}{2*16}=\frac{-16}{32}
=-1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+32}{2*16}=\frac{48}{32}
=1+1/2
$