-6t^2-18t+16=0

Solution for -6t^2-18t+16=0 equation:

-6t^2-18t+16=0

a = -6; b = -18; c = +16;
Δ = b2-4ac
Δ = -182-4·(-6)·16
Δ = 708
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:

$\sqrt{\Delta}=\sqrt{708}=\sqrt{4*177}=\sqrt{4}*\sqrt{177}=2\sqrt{177}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2\sqrt{177}}{2*-6}=\frac{18-2\sqrt{177}}{-12}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2\sqrt{177}}{2*-6}=\frac{18+2\sqrt{177}}{-12}
$