0=16-18t^2

Solution for 0=16-18t^2 equation:

0=16-18t^2

We move all terms to the left:

0-(16-18t^2)=0

We add all the numbers together, and all the variables

-(16-18t^2)=0

We get rid of parentheses

18t^2-16=0

a = 18; b = 0; c = -16;
Δ = b2-4ac
Δ = 02-4·18·(-16)
Δ = 1152
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:

$\sqrt{\Delta}=\sqrt{1152}=\sqrt{576*2}=\sqrt{576}*\sqrt{2}=24\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-24\sqrt{2}}{2*18}=\frac{0-24\sqrt{2}}{36}
=-\frac{24\sqrt{2}}{36}
=-\frac{2\sqrt{2}}{3}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+24\sqrt{2}}{2*18}=\frac{0+24\sqrt{2}}{36}
=\frac{24\sqrt{2}}{36}
=\frac{2\sqrt{2}}{3}
$