4(3x^2-4)=0

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

4(3x^2-4)=0

We multiply parentheses

12x^2-16=0

a = 12; b = 0; c = -16;
Δ = b2-4ac
Δ = 02-4·12·(-16)
Δ = 768
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:

$\sqrt{\Delta}=\sqrt{768}=\sqrt{256*3}=\sqrt{256}*\sqrt{3}=16\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{3}}{2*12}=\frac{0-16\sqrt{3}}{24}
=-\frac{16\sqrt{3}}{24}
=-\frac{2\sqrt{3}}{3}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{3}}{2*12}=\frac{0+16\sqrt{3}}{24}
=\frac{16\sqrt{3}}{24}
=\frac{2\sqrt{3}}{3}
$