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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

-(3(+8x^2+2x+4x+1))=0


We calculate terms in parentheses: -(3(+8x^2+2x+4x+1)), so:

3(+8x^2+2x+4x+1)

We multiply parentheses

24x^2+6x+12x+3

We add all the numbers together, and all the variables

24x^2+18x+3

Back to the equation:

-(24x^2+18x+3)


We get rid of parentheses

-24x^2-18x-3=0

a = -24; b = -18; c = -3;
Δ = b2-4ac
Δ = -182-4·(-24)·(-3)
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-6}{2*-24}=\frac{12}{-48}
=-1/4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+6}{2*-24}=\frac{24}{-48}
=-1/2
$