2-3t^2=0

Solution for 2-3t^2=0 equation:

2-3t^2=0

a = -3; b = 0; c = +2;
Δ = b2-4ac
Δ = 02-4·(-3)·2
Δ = 24
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:

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