17=-14t+4.9t^2

Solution for 17=-14t+4.9t^2 equation:

17=-14t+4.9t^2

We move all terms to the left:

17-(-14t+4.9t^2)=0

We get rid of parentheses

-4.9t^2+14t+17=0

a = -4.9; b = 14; c = +17;
Δ = b2-4ac
Δ = 142-4·(-4.9)·17
Δ = 529.2
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{-(14)-\sqrt{529.2}}{2*-4.9}=\frac{-14-\sqrt{529.2}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+\sqrt{529.2}}{2*-4.9}=\frac{-14+\sqrt{529.2}}{-9.8}
$