0=16t^2-20t+2

Solution for 0=16t^2-20t+2 equation:

0=16t^2-20t+2

We move all terms to the left:

0-(16t^2-20t+2)=0

We add all the numbers together, and all the variables

-(16t^2-20t+2)=0

We get rid of parentheses

-16t^2+20t-2=0

a = -16; b = 20; c = -2;
Δ = b2-4ac
Δ = 202-4·(-16)·(-2)
Δ = 272
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:

$\sqrt{\Delta}=\sqrt{272}=\sqrt{16*17}=\sqrt{16}*\sqrt{17}=4\sqrt{17}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{17}}{2*-16}=\frac{-20-4\sqrt{17}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{17}}{2*-16}=\frac{-20+4\sqrt{17}}{-32}
$