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 = 139.8
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{139.8}}{2*4.9}=\frac{-9-\sqrt{139.8}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+\sqrt{139.8}}{2*4.9}=\frac{-9+\sqrt{139.8}}{9.8}
$