-4.9t^2+14t-6.4=0

Solution for -4.9t^2+14t-6.4=0 equation:

-4.9t^2+14t-6.4=0

a = -4.9; b = 14; c = -6.4;
Δ = b2-4ac
Δ = 142-4·(-4.9)·(-6.4)
Δ = 70.56
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{-(14)-\sqrt{70.56}}{2*-4.9}=\frac{-14-\sqrt{70.56}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+\sqrt{70.56}}{2*-4.9}=\frac{-14+\sqrt{70.56}}{-9.8}
$