0=16t^2+11t+0

Solution for 0=16t^2+11t+0 equation:

0=16t^2+11t+0

We move all terms to the left:

0-(16t^2+11t+0)=0

We add all the numbers together, and all the variables

-(16t^2+11t+0)=0

We get rid of parentheses

-16t^2-11t-0=0

We add all the numbers together, and all the variables

-16t^2-11t=0

a = -16; b = -11; c = 0;
Δ = b2-4ac
Δ = -112-4·(-16)·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*-16}=\frac{0}{-32}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+11}{2*-16}=\frac{22}{-32}
=-11/16
$