0=-6t2+48t

Solution for 0=-6t2+48t equation:

0=-6t^2+48t

We move all terms to the left:

0-(-6t^2+48t)=0

We add all the numbers together, and all the variables

-(-6t^2+48t)=0

We get rid of parentheses

6t^2-48t=0

a = 6; b = -48; c = 0;
Δ = b2-4ac
Δ = -482-4·6·0
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*6}=\frac{0}{12}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+48}{2*6}=\frac{96}{12}
=8
$