4.9t^2+11t-41=0

Solution for 4.9t^2+11t-41=0 equation:

4.9t^2+11t-41=0

a = 4.9; b = 11; c = -41;
Δ = b2-4ac
Δ = 112-4·4.9·(-41)
Δ = 924.6
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{924.6}}{2*4.9}=\frac{-11-\sqrt{924.6}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+\sqrt{924.6}}{2*4.9}=\frac{-11+\sqrt{924.6}}{9.8}
$