-4.9t2-13t+4=0

Solution for -4.9t2-13t+4=0 equation:

-4.9t^2-13t+4=0

a = -4.9; b = -13; c = +4;
Δ = b2-4ac
Δ = -132-4·(-4.9)·4
Δ = 247.4
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{-(-13)-\sqrt{247.4}}{2*-4.9}=\frac{13-\sqrt{247.4}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+\sqrt{247.4}}{2*-4.9}=\frac{13+\sqrt{247.4}}{-9.8}
$