4.9t^2+9t-450=0

Solution for 4.9t^2+9t-450=0 equation:

4.9t^2+9t-450=0

a = 4.9; b = 9; c = -450;
Δ = b2-4ac
Δ = 92-4·4.9·(-450)
Δ = 8901
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{8901}=\sqrt{9*989}=\sqrt{9}*\sqrt{989}=3\sqrt{989}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-3\sqrt{989}}{2*4.9}=\frac{-9-3\sqrt{989}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+3\sqrt{989}}{2*4.9}=\frac{-9+3\sqrt{989}}{9.8}
$