0=-t^2+8t

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

0=-t^2+8t

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

t^2-8t=0

a = 1; b = -8; c = 0;
Δ = b2-4ac
Δ = -82-4·1·0
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-8}{2*1}=\frac{0}{2}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+8}{2*1}=\frac{16}{2}
=8
$