4.9t^2-5t-0.5=0

Solution for 4.9t^2-5t-0.5=0 equation:

4.9t^2-5t-0.5=0

a = 4.9; b = -5; c = -0.5;
Δ = b2-4ac
Δ = -52-4·4.9·(-0.5)
Δ = 34.8
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{-(-5)-\sqrt{34.8}}{2*4.9}=\frac{5-\sqrt{34.8}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{34.8}}{2*4.9}=\frac{5+\sqrt{34.8}}{9.8}
$