-0.2t^2+2t=0

Solution for -0.2t^2+2t=0 equation:

-0.2t^2+2t=0

a = -0.2; b = 2; c = 0;
Δ = b2-4ac
Δ = 22-4·(-0.2)·0
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2}{2*-0.2}=\frac{-4}{-0.4}
=+10
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2}{2*-0.2}=\frac{0}{-0.4}
=0
$