-4.9t^2+900t+2.4=0

Solution for -4.9t^2+900t+2.4=0 equation:

-4.9t^2+900t+2.4=0

a = -4.9; b = 900; c = +2.4;
Δ = b2-4ac
Δ = 9002-4·(-4.9)·2.4
Δ = 810047.04
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{-(900)-\sqrt{810047.04}}{2*-4.9}=\frac{-900-\sqrt{810047.04}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(900)+\sqrt{810047.04}}{2*-4.9}=\frac{-900+\sqrt{810047.04}}{-9.8}
$