0=-16t^2+20t-5.95

Solution for 0=-16t^2+20t-5.95 equation:

0=-16t^2+20t-5.95

We move all terms to the left:

0-(-16t^2+20t-5.95)=0

We add all the numbers together, and all the variables

-(-16t^2+20t-5.95)=0

We get rid of parentheses

16t^2-20t+5.95=0

a = 16; b = -20; c = +5.95;
Δ = b2-4ac
Δ = -202-4·16·5.95
Δ = 19.2
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{-(-20)-\sqrt{19.2}}{2*16}=\frac{20-\sqrt{19.2}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+\sqrt{19.2}}{2*16}=\frac{20+\sqrt{19.2}}{32}
$