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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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


We calculate terms in parentheses: -(-0.5t(t-8)), so:

-0.5t(t-8)

We multiply parentheses

-0t^2-0t

We add all the numbers together, and all the variables

-1t^2-1t

Back to the equation:

-(-1t^2-1t)


We get rid of parentheses

1t^2+1t=0

We add all the numbers together, and all the variables

t^2+t=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{-2}{2}
=-1
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*1}=\frac{0}{2}
=0
$