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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

16t^2-1t=0

a = 16; b = -1; c = 0;
Δ = b2-4ac
Δ = -12-4·16·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-1}{2*16}=\frac{0}{32}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+1}{2*16}=\frac{2}{32}
=1/16
$