0=16t^2+48t+3.5

Solution for 0=16t^2+48t+3.5 equation:

0=16t^2+48t+3.5

We move all terms to the left:

0-(16t^2+48t+3.5)=0

We add all the numbers together, and all the variables

-(16t^2+48t+3.5)=0

We get rid of parentheses

-16t^2-48t-3.5=0

a = -16; b = -48; c = -3.5;
Δ = b2-4ac
Δ = -482-4·(-16)·(-3.5)
Δ = 2080
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:

$\sqrt{\Delta}=\sqrt{2080}=\sqrt{16*130}=\sqrt{16}*\sqrt{130}=4\sqrt{130}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-4\sqrt{130}}{2*-16}=\frac{48-4\sqrt{130}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+4\sqrt{130}}{2*-16}=\frac{48+4\sqrt{130}}{-32}
$