4.9t^2+4.96t-2.3=0

Solution for 4.9t^2+4.96t-2.3=0 equation:

4.9t^2+4.96t-2.3=0

a = 4.9; b = 4.96; c = -2.3;
Δ = b2-4ac
Δ = 4.962-4·4.9·(-2.3)
Δ = 69.6816
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{-(4.96)-\sqrt{69.6816}}{2*4.9}=\frac{-4.96-\sqrt{69.6816}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4.96)+\sqrt{69.6816}}{2*4.9}=\frac{-4.96+\sqrt{69.6816}}{9.8}
$