-10t^2-5t=0

Solution for -10t^2-5t=0 equation:

-10t^2-5t=0

a = -10; b = -5; c = 0;
Δ = b2-4ac
Δ = -52-4·(-10)·0
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-5}{2*-10}=\frac{0}{-20}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+5}{2*-10}=\frac{10}{-20}
=-1/2
$