0=-4.9t^2+20t-16

Solution for 0=-4.9t^2+20t-16 equation:

0=-4.9t^2+20t-16

We move all terms to the left:

0-(-4.9t^2+20t-16)=0

We add all the numbers together, and all the variables

-(-4.9t^2+20t-16)=0

We get rid of parentheses

4.9t^2-20t+16=0

a = 4.9; b = -20; c = +16;
Δ = b2-4ac
Δ = -202-4·4.9·16
Δ = 86.4
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{86.4}}{2*4.9}=\frac{20-\sqrt{86.4}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+\sqrt{86.4}}{2*4.9}=\frac{20+\sqrt{86.4}}{9.8}
$