121+41.8t-4.9t^2=0

Solution for 121+41.8t-4.9t^2=0 equation:

121+41.8t-4.9t^2=0

a = -4.9; b = 41.8; c = +121;
Δ = b2-4ac
Δ = 41.82-4·(-4.9)·121
Δ = 4118.84
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{-(41.8)-\sqrt{4118.84}}{2*-4.9}=\frac{-41.8-\sqrt{4118.84}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(41.8)+\sqrt{4118.84}}{2*-4.9}=\frac{-41.8+\sqrt{4118.84}}{-9.8}
$