0=16t^2+48t+4.

Solution for 0=16t^2+48t+4. equation:

0=16t^2+48t+4.

We move all terms to the left:

0-(16t^2+48t+4.)=0

We add all the numbers together, and all the variables

-(16t^2+48t+4.)=0

We get rid of parentheses

-16t^2-48t-4.=0

We add all the numbers together, and all the variables

-16t^2-48t-4=0

a = -16; b = -48; c = -4;
Δ = b2-4ac
Δ = -482-4·(-16)·(-4)
Δ = 2048
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:

$\sqrt{\Delta}=\sqrt{2048}=\sqrt{1024*2}=\sqrt{1024}*\sqrt{2}=32\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-32\sqrt{2}}{2*-16}=\frac{48-32\sqrt{2}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+32\sqrt{2}}{2*-16}=\frac{48+32\sqrt{2}}{-32}
$