-4.9t^2+11t=0

Solution for -4.9t^2+11t=0 equation:

-4.9t^2+11t=0

a = -4.9; b = 11; c = 0;
Δ = b2-4ac
Δ = 112-4·(-4.9)·0
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-11}{2*-4.9}=\frac{-22}{-9.8}
=2+2.4/9.8
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+11}{2*-4.9}=\frac{0}{-9.8}
=0
$