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}
$