17=-16t^2+28t+7

Solution for 17=-16t^2+28t+7 equation:

17=-16t^2+28t+7

We move all terms to the left:

17-(-16t^2+28t+7)=0

We get rid of parentheses

16t^2-28t-7+17=0

We add all the numbers together, and all the variables

16t^2-28t+10=0

a = 16; b = -28; c = +10;
Δ = b2-4ac
Δ = -282-4·16·10
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-12}{2*16}=\frac{16}{32}
=1/2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+12}{2*16}=\frac{40}{32}
=1+1/4
$