1/3r^2-4=0

Solution for 1/3r^2-4=0 equation:

1/3r^2-4=0


Domain of the equation: 3r^2!=0

r^2!=0/3

r^2!=√0

r!=0

r∈R


We multiply all the terms by the denominator


-4*3r^2+1=0

Wy multiply elements

-12r^2+1=0

a = -12; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·(-12)·1
Δ = 48
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{3}}{2*-12}=\frac{0-4\sqrt{3}}{-24}
=-\frac{4\sqrt{3}}{-24}
=-\frac{\sqrt{3}}{-6}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{3}}{2*-12}=\frac{0+4\sqrt{3}}{-24}
=\frac{4\sqrt{3}}{-24}
=\frac{\sqrt{3}}{-6}
$