4.9t^2+4t-3=0

Solution for 4.9t^2+4t-3=0 equation:

4.9t^2+4t-3=0

a = 4.9; b = 4; c = -3;
Δ = b2-4ac
Δ = 42-4·4.9·(-3)
Δ = 74.8
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{-(4)-\sqrt{74.8}}{2*4.9}=\frac{-4-\sqrt{74.8}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+\sqrt{74.8}}{2*4.9}=\frac{-4+\sqrt{74.8}}{9.8}
$