0=-16t^2+169t+46

Solution for 0=-16t^2+169t+46 equation:

0=-16t^2+169t+46

We move all terms to the left:

0-(-16t^2+169t+46)=0

We add all the numbers together, and all the variables

-(-16t^2+169t+46)=0

We get rid of parentheses

16t^2-169t-46=0

a = 16; b = -169; c = -46;
Δ = b2-4ac
Δ = -1692-4·16·(-46)
Δ = 31505
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-169)-\sqrt{31505}}{2*16}=\frac{169-\sqrt{31505}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-169)+\sqrt{31505}}{2*16}=\frac{169+\sqrt{31505}}{32}
$