17=4+30t-16t^2

Solution for 17=4+30t-16t^2 equation:

17=4+30t-16t^2

We move all terms to the left:

17-(4+30t-16t^2)=0

We get rid of parentheses

16t^2-30t-4+17=0

We add all the numbers together, and all the variables

16t^2-30t+13=0

a = 16; b = -30; c = +13;
Δ = b2-4ac
Δ = -302-4·16·13
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{-(-30)-2\sqrt{17}}{2*16}=\frac{30-2\sqrt{17}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-30)+2\sqrt{17}}{2*16}=\frac{30+2\sqrt{17}}{32}
$