0=-16t^2+169

Solution for 0=-16t^2+169 equation:

0=-16t^2+169

We move all terms to the left:

0-(-16t^2+169)=0

We add all the numbers together, and all the variables

-(-16t^2+169)=0

We get rid of parentheses

16t^2-169=0

a = 16; b = 0; c = -169;
Δ = b2-4ac
Δ = 02-4·16·(-169)
Δ = 10816
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{10816}=104$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-104}{2*16}=\frac{-104}{32}
=-3+1/4
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+104}{2*16}=\frac{104}{32}
=3+1/4
$