0=-16t^2+22

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

0=-16t^2+22

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

16t^2-22=0

a = 16; b = 0; c = -22;
Δ = b2-4ac
Δ = 02-4·16·(-22)
Δ = 1408
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:

$\sqrt{\Delta}=\sqrt{1408}=\sqrt{64*22}=\sqrt{64}*\sqrt{22}=8\sqrt{22}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{22}}{2*16}=\frac{0-8\sqrt{22}}{32}
=-\frac{8\sqrt{22}}{32}
=-\frac{\sqrt{22}}{4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{22}}{2*16}=\frac{0+8\sqrt{22}}{32}
=\frac{8\sqrt{22}}{32}
=\frac{\sqrt{22}}{4}
$