(((x^2)-2)/4)-(((3x^2)+4)/8)-((2x+3)/6)-((6x+4)/12)=0

Solution for (((x^2)-2)/4)-(((3x^2)+4)/8)-((2x+3)/6)-((6x+4)/12)=0 equation:

(((x^2)-2)/4)-(((3x^2)+4)/8)-((2x+3)/6)-((6x+4)/12)=0

We calculate fractions


(-36x^2)/()+12x^2/()+(-24x)/()+(-36x)/()=0

We multiply all the terms by the denominator


(-36x^2)+12x^2+(-24x)+(-36x)=0

We add all the numbers together, and all the variables

12x^2+(-36x^2)+(-24x)+(-36x)=0

We get rid of parentheses

12x^2-36x^2-24x-36x=0

We add all the numbers together, and all the variables

-24x^2-60x=0

a = -24; b = -60; c = 0;
Δ = b2-4ac
Δ = -602-4·(-24)·0
Δ = 3600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3600}=60$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-60)-60}{2*-24}=\frac{0}{-48}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-60)+60}{2*-24}=\frac{120}{-48}
=-2+1/2
$