-6=-16t^2+27t

Solution for -6=-16t^2+27t equation:

-6=-16t^2+27t

We move all terms to the left:

-6-(-16t^2+27t)=0

We get rid of parentheses

16t^2-27t-6=0

a = 16; b = -27; c = -6;
Δ = b2-4ac
Δ = -272-4·16·(-6)
Δ = 1113
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-27)-\sqrt{1113}}{2*16}=\frac{27-\sqrt{1113}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-27)+\sqrt{1113}}{2*16}=\frac{27+\sqrt{1113}}{32}
$