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

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

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


Domain of the equation: (x^2+4)^3!=0

x∈R


We multiply all the terms by the denominator


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


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

-6(3x^2-4)

We multiply parentheses

-18x^2+24

Back to the equation:

+(-18x^2+24)


We get rid of parentheses

-18x^2+24=0

a = -18; b = 0; c = +24;
Δ = b2-4ac
Δ = 02-4·(-18)·24
Δ = 1728
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1728}=\sqrt{576*3}=\sqrt{576}*\sqrt{3}=24\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-24\sqrt{3}}{2*-18}=\frac{0-24\sqrt{3}}{-36}
=-\frac{24\sqrt{3}}{-36}
=-\frac{2\sqrt{3}}{-3}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+24\sqrt{3}}{2*-18}=\frac{0+24\sqrt{3}}{-36}
=\frac{24\sqrt{3}}{-36}
=\frac{2\sqrt{3}}{-3}
$