-4.9t^2+11t+55=0

Solution for -4.9t^2+11t+55=0 equation:

-4.9t^2+11t+55=0

a = -4.9; b = 11; c = +55;
Δ = b2-4ac
Δ = 112-4·(-4.9)·55
Δ = 1199
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{1199}}{2*-4.9}=\frac{-11-\sqrt{1199}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+\sqrt{1199}}{2*-4.9}=\frac{-11+\sqrt{1199}}{-9.8}
$