0=4.9t^2-20t+11

Solution for 0=4.9t^2-20t+11 equation:

0=4.9t^2-20t+11

We move all terms to the left:

0-(4.9t^2-20t+11)=0

We add all the numbers together, and all the variables

-(4.9t^2-20t+11)=0

We get rid of parentheses

-4.9t^2+20t-11=0

a = -4.9; b = 20; c = -11;
Δ = b2-4ac
Δ = 202-4·(-4.9)·(-11)
Δ = 184.4
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{184.4}}{2*-4.9}=\frac{-20-\sqrt{184.4}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+\sqrt{184.4}}{2*-4.9}=\frac{-20+\sqrt{184.4}}{-9.8}
$