0=48t-8t^2

Solution for 0=48t-8t^2 equation:

0=48t-8t^2

We move all terms to the left:

0-(48t-8t^2)=0

We add all the numbers together, and all the variables

-(48t-8t^2)=0

We get rid of parentheses

8t^2-48t=0

a = 8; b = -48; c = 0;
Δ = b2-4ac
Δ = -482-4·8·0
Δ = 2304
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2304}=48$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-48}{2*8}=\frac{0}{16}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+48}{2*8}=\frac{96}{16}
=6
$