4.9t^2-8t-13=0

Solution for 4.9t^2-8t-13=0 equation:

4.9t^2-8t-13=0

a = 4.9; b = -8; c = -13;
Δ = b2-4ac
Δ = -82-4·4.9·(-13)
Δ = 318.8
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{-(-8)-\sqrt{318.8}}{2*4.9}=\frac{8-\sqrt{318.8}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+\sqrt{318.8}}{2*4.9}=\frac{8+\sqrt{318.8}}{9.8}
$