4.9t^2-11t-2.1=0

Solution for 4.9t^2-11t-2.1=0 equation:

4.9t^2-11t-2.1=0

a = 4.9; b = -11; c = -2.1;
Δ = b2-4ac
Δ = -112-4·4.9·(-2.1)
Δ = 162.16
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{162.16}}{2*4.9}=\frac{11-\sqrt{162.16}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+\sqrt{162.16}}{2*4.9}=\frac{11+\sqrt{162.16}}{9.8}
$