0=-16t^2+20t-6

Solution for 0=-16t^2+20t-6 equation:

0=-16t^2+20t-6

We move all terms to the left:

0-(-16t^2+20t-6)=0

We add all the numbers together, and all the variables

-(-16t^2+20t-6)=0

We get rid of parentheses

16t^2-20t+6=0

a = 16; b = -20; c = +6;
Δ = b2-4ac
Δ = -202-4·16·6
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4}{2*16}=\frac{16}{32}
=1/2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4}{2*16}=\frac{24}{32}
=3/4
$