0=-16t^2+48t+28

Solution for 0=-16t^2+48t+28 equation:

0=-16t^2+48t+28

We move all terms to the left:

0-(-16t^2+48t+28)=0

We add all the numbers together, and all the variables

-(-16t^2+48t+28)=0

We get rid of parentheses

16t^2-48t-28=0

a = 16; b = -48; c = -28;
Δ = b2-4ac
Δ = -482-4·16·(-28)
Δ = 4096
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4096}=64$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-64}{2*16}=\frac{-16}{32}
=-1/2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+64}{2*16}=\frac{112}{32}
=3+1/2
$