0=-16t^2+22t

Solution for 0=-16t^2+22t equation:

0=-16t^2+22t

We move all terms to the left:

0-(-16t^2+22t)=0

We add all the numbers together, and all the variables

-(-16t^2+22t)=0

We get rid of parentheses

16t^2-22t=0

a = 16; b = -22; c = 0;
Δ = b2-4ac
Δ = -222-4·16·0
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-22}{2*16}=\frac{0}{32}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+22}{2*16}=\frac{44}{32}
=1+3/8
$