1.5-6t^2+6t=0

Solution for 1.5-6t^2+6t=0 equation:

1.5-6t^2+6t=0

a = -6; b = 6; c = +1.5;
Δ = b2-4ac
Δ = 62-4·(-6)·1.5
Δ = 72
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:

$\sqrt{\Delta}=\sqrt{72}=\sqrt{36*2}=\sqrt{36}*\sqrt{2}=6\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6\sqrt{2}}{2*-6}=\frac{-6-6\sqrt{2}}{-12}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6\sqrt{2}}{2*-6}=\frac{-6+6\sqrt{2}}{-12}
$