-4.9t^2+9t-3=0

Solution for -4.9t^2+9t-3=0 equation:

-4.9t^2+9t-3=0

a = -4.9; b = 9; c = -3;
Δ = b2-4ac
Δ = 92-4·(-4.9)·(-3)
Δ = 22.2
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{22.2}}{2*-4.9}=\frac{-9-\sqrt{22.2}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+\sqrt{22.2}}{2*-4.9}=\frac{-9+\sqrt{22.2}}{-9.8}
$