-0.1t^2+t=0

Solution for -0.1t^2+t=0 equation:

-0.1t^2+t=0

a = -0.1; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·(-0.1)·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*-0.1}=\frac{-2}{-0.2}
=+10
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*-0.1}=\frac{0}{-0.2}
=0
$