0=-16t^2+112t+48

Solution for 0=-16t^2+112t+48 equation:

0=-16t^2+112t+48

We move all terms to the left:

0-(-16t^2+112t+48)=0

We add all the numbers together, and all the variables

-(-16t^2+112t+48)=0

We get rid of parentheses

16t^2-112t-48=0

a = 16; b = -112; c = -48;
Δ = b2-4ac
Δ = -1122-4·16·(-48)
Δ = 15616
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:

$\sqrt{\Delta}=\sqrt{15616}=\sqrt{256*61}=\sqrt{256}*\sqrt{61}=16\sqrt{61}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-112)-16\sqrt{61}}{2*16}=\frac{112-16\sqrt{61}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-112)+16\sqrt{61}}{2*16}=\frac{112+16\sqrt{61}}{32}
$