12t^2-5=0

Solution for 12t^2-5=0 equation:

12t^2-5=0

a = 12; b = 0; c = -5;
Δ = b2-4ac
Δ = 02-4·12·(-5)
Δ = 240
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:

$\sqrt{\Delta}=\sqrt{240}=\sqrt{16*15}=\sqrt{16}*\sqrt{15}=4\sqrt{15}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{15}}{2*12}=\frac{0-4\sqrt{15}}{24}
=-\frac{4\sqrt{15}}{24}
=-\frac{\sqrt{15}}{6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{15}}{2*12}=\frac{0+4\sqrt{15}}{24}
=\frac{4\sqrt{15}}{24}
=\frac{\sqrt{15}}{6}
$