What you have is the quadratic - 4.9 t^2 + 82*sin(radians(35)) t + 3300 = 0. Look up the solution equation on the web. A = -4.9, B = 82*sin(radians(35)), C = 3300 ans solve for t.
But better yet, use a quadratic equation calculator (there are plenty on the web) and plug in the coefficients for the answers (there will be two).
Answers & Comments
What you have is the quadratic - 4.9 t^2 + 82*sin(radians(35)) t + 3300 = 0. Look up the solution equation on the web. A = -4.9, B = 82*sin(radians(35)), C = 3300 ans solve for t.
But better yet, use a quadratic equation calculator (there are plenty on the web) and plug in the coefficients for the answers (there will be two).
A = -4.900
B = 47.033
C = 3300.000
Large Solution: -21.592
Small Solution: 31.191
Leaving out units for neatness:
½(-9.81)t² + 82sin(35°)t - (-3300) = 0
Replace ½(-9.81) by -4.905
Replace 82sin(35°) by 47.03
Replace - (-3300 by + 3300
giving:
-4.905t² + 47.03t + 3300 = 0
This is a standard quadratic equation like ax² + bx + c = 0, but using 't' instead of 'x'.
a = -4.905
b = 47.03
c = 3300
The solutions are given by the standard equation:
t = [-b ± √(b² - 4ac)]/(2a)
The arithmetic is too messy to type out. I get:
t = 31.2s or t = -21.6s
(If this is for a projectile problem where the projectile starts at t=0, we ignore the negative solution as it is not meaningful.)
t = 31.17s