-4.9t^2-9t+150=0

Solution for -4.9t^2-9t+150=0 equation:

-4.9t^2-9t+150=0

a = -4.9; b = -9; c = +150;
Δ = b2-4ac
Δ = -92-4·(-4.9)·150
Δ = 3021
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)-\sqrt{3021}}{2*-4.9}=\frac{9-\sqrt{3021}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+\sqrt{3021}}{2*-4.9}=\frac{9+\sqrt{3021}}{-9.8}
$