0=-16t^2+32t-5

Solution for 0=-16t^2+32t-5 equation:

0=-16t^2+32t-5

We move all terms to the left:

0-(-16t^2+32t-5)=0

We add all the numbers together, and all the variables

-(-16t^2+32t-5)=0

We get rid of parentheses

16t^2-32t+5=0

a = 16; b = -32; c = +5;
Δ = b2-4ac
Δ = -322-4·16·5
Δ = 704
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:

$\sqrt{\Delta}=\sqrt{704}=\sqrt{64*11}=\sqrt{64}*\sqrt{11}=8\sqrt{11}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-8\sqrt{11}}{2*16}=\frac{32-8\sqrt{11}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+8\sqrt{11}}{2*16}=\frac{32+8\sqrt{11}}{32}
$