-0.105t^2+-0.5t=0

Solution for -0.105t^2+-0.5t=0 equation:

-0.105t^2+-0.5t=0

We add all the numbers together, and all the variables

-0.105t^2-0.5t=0

a = -0.105; b = -0.5; c = 0;
Δ = b2-4ac
Δ = -0.52-4·(-0.105)·0
Δ = 0.25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-0.5)-\sqrt{0.25}}{2*-0.105}=\frac{0.5-\sqrt{0.25}}{-0.21}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-0.5)+\sqrt{0.25}}{2*-0.105}=\frac{0.5+\sqrt{0.25}}{-0.21}
$