1/6t^2-4=0

Solution for 1/6t^2-4=0 equation:

1/6t^2-4=0


Domain of the equation: 6t^2!=0

t^2!=0/6

t^2!=√0

t!=0

t∈R


We multiply all the terms by the denominator


-4*6t^2+1=0

Wy multiply elements

-24t^2+1=0

a = -24; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·(-24)·1
Δ = 96
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{6}}{2*-24}=\frac{0-4\sqrt{6}}{-48}
=-\frac{4\sqrt{6}}{-48}
=-\frac{\sqrt{6}}{-12}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{6}}{2*-24}=\frac{0+4\sqrt{6}}{-48}
=\frac{4\sqrt{6}}{-48}
=\frac{\sqrt{6}}{-12}
$