0=16t^2-17t+4

Solution for 0=16t^2-17t+4 equation:

0=16t^2-17t+4

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

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

a = -16; b = 17; c = -4;
Δ = b2-4ac
Δ = 172-4·(-16)·(-4)
Δ = 33
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{-(17)-\sqrt{33}}{2*-16}=\frac{-17-\sqrt{33}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+\sqrt{33}}{2*-16}=\frac{-17+\sqrt{33}}{-32}
$