17=5t+29t-16t^2

Solution for 17=5t+29t-16t^2 equation:

17=5t+29t-16t^2

We move all terms to the left:

17-(5t+29t-16t^2)=0

We get rid of parentheses

16t^2-5t-29t+17=0

We add all the numbers together, and all the variables

16t^2-34t+17=0

a = 16; b = -34; c = +17;
Δ = b2-4ac
Δ = -342-4·16·17
Δ = 68
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:

$\sqrt{\Delta}=\sqrt{68}=\sqrt{4*17}=\sqrt{4}*\sqrt{17}=2\sqrt{17}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-34)-2\sqrt{17}}{2*16}=\frac{34-2\sqrt{17}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-34)+2\sqrt{17}}{2*16}=\frac{34+2\sqrt{17}}{32}
$