22=16t^2

Solution for 22=16t^2 equation:

22=16t^2

We move all terms to the left:

22-(16t^2)=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}
$