-6=-16t^2+48t

Solution for -6=-16t^2+48t equation:

-6=-16t^2+48t

We move all terms to the left:

-6-(-16t^2+48t)=0

We get rid of parentheses

16t^2-48t-6=0

a = 16; b = -48; c = -6;
Δ = b2-4ac
Δ = -482-4·16·(-6)
Δ = 2688
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:

$\sqrt{\Delta}=\sqrt{2688}=\sqrt{64*42}=\sqrt{64}*\sqrt{42}=8\sqrt{42}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-8\sqrt{42}}{2*16}=\frac{48-8\sqrt{42}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+8\sqrt{42}}{2*16}=\frac{48+8\sqrt{42}}{32}
$