-4.4t^2+10t=0

Solution for -4.4t^2+10t=0 equation:

-4.4t^2+10t=0

a = -4.4; b = 10; c = 0;
Δ = b2-4ac
Δ = 102-4·(-4.4)·0
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-10}{2*-4.4}=\frac{-20}{-8.8}
=2+1/3.66666666667
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+10}{2*-4.4}=\frac{0}{-8.8}
=0
$