0=4.9t^2+11t-0.4

Solution for 0=4.9t^2+11t-0.4 equation:

0=4.9t^2+11t-0.4

We move all terms to the left:

0-(4.9t^2+11t-0.4)=0

We add all the numbers together, and all the variables

-(4.9t^2+11t-0.4)=0

We get rid of parentheses

-4.9t^2-11t+0.4=0

a = -4.9; b = -11; c = +0.4;
Δ = b2-4ac
Δ = -112-4·(-4.9)·0.4
Δ = 128.84
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{-(-11)-\sqrt{128.84}}{2*-4.9}=\frac{11-\sqrt{128.84}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+\sqrt{128.84}}{2*-4.9}=\frac{11+\sqrt{128.84}}{-9.8}
$