-4.9t^2+9.8t+150=0

Solution for -4.9t^2+9.8t+150=0 equation:

-4.9t^2+9.8t+150=0

a = -4.9; b = 9.8; c = +150;
Δ = b2-4ac
Δ = 9.82-4·(-4.9)·150
Δ = 3036.04
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{-(9.8)-\sqrt{3036.04}}{2*-4.9}=\frac{-9.8-\sqrt{3036.04}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9.8)+\sqrt{3036.04}}{2*-4.9}=\frac{-9.8+\sqrt{3036.04}}{-9.8}
$