0=16t^2+48t+6

Solution for 0=16t^2+48t+6 equation:

0=16t^2+48t+6

We move all terms to the left:

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

We add all the numbers together, and all the variables

-(16t^2+48t+6)=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 = 1920
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:

$\sqrt{\Delta}=\sqrt{1920}=\sqrt{64*30}=\sqrt{64}*\sqrt{30}=8\sqrt{30}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-8\sqrt{30}}{2*-16}=\frac{48-8\sqrt{30}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+8\sqrt{30}}{2*-16}=\frac{48+8\sqrt{30}}{-32}
$