0=-16t^2+189

Solution for 0=-16t^2+189 equation:

0=-16t^2+189

We move all terms to the left:

0-(-16t^2+189)=0

We add all the numbers together, and all the variables

-(-16t^2+189)=0

We get rid of parentheses

16t^2-189=0

a = 16; b = 0; c = -189;
Δ = b2-4ac
Δ = 02-4·16·(-189)
Δ = 12096
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:

$\sqrt{\Delta}=\sqrt{12096}=\sqrt{576*21}=\sqrt{576}*\sqrt{21}=24\sqrt{21}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-24\sqrt{21}}{2*16}=\frac{0-24\sqrt{21}}{32}
=-\frac{24\sqrt{21}}{32}
=-\frac{3\sqrt{21}}{4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+24\sqrt{21}}{2*16}=\frac{0+24\sqrt{21}}{32}
=\frac{24\sqrt{21}}{32}
=\frac{3\sqrt{21}}{4}
$