4.9t^2-(16.42)t-11=0

Solution for 4.9t^2-(16.42)t-11=0 equation:

4.9t^2-(16.42)t-11=0

We multiply parentheses

4.9t^2-16.42t-11=0

a = 4.9; b = -16.42; c = -11;
Δ = b2-4ac
Δ = -16.422-4·4.9·(-11)
Δ = 485.2164
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{-(-16.42)-\sqrt{485.2164}}{2*4.9}=\frac{16.42-\sqrt{485.2164}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16.42)+\sqrt{485.2164}}{2*4.9}=\frac{16.42+\sqrt{485.2164}}{9.8}
$