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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

16t^2-22t-2=0

a = 16; b = -22; c = -2;
Δ = b2-4ac
Δ = -222-4·16·(-2)
Δ = 612
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{612}=\sqrt{36*17}=\sqrt{36}*\sqrt{17}=6\sqrt{17}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-6\sqrt{17}}{2*16}=\frac{22-6\sqrt{17}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+6\sqrt{17}}{2*16}=\frac{22+6\sqrt{17}}{32}
$