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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

16t^2-0t-48=0

We add all the numbers together, and all the variables

16t^2-1t-48=0

a = 16; b = -1; c = -48;
Δ = b2-4ac
Δ = -12-4·16·(-48)
Δ = 3073
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-\sqrt{3073}}{2*16}=\frac{1-\sqrt{3073}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{3073}}{2*16}=\frac{1+\sqrt{3073}}{32}
$