-0.5(t)(t-8)=0

Solution for -0.5(t)(t-8)=0 equation:

-0.5(t)(t-8)=0

We multiply parentheses

-0t^2-0t=0

We add all the numbers together, and all the variables

-1t^2-1t=0

a = -1; b = -1; c = 0;
Δ = b2-4ac
Δ = -12-4·(-1)·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*-1}=\frac{0}{-2}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+1}{2*-1}=\frac{2}{-2}
=-1
$