4.905t^2-10t-30=0

Solution for 4.905t^2-10t-30=0 equation:

4.905t^2-10t-30=0

a = 4.905; b = -10; c = -30;
Δ = b2-4ac
Δ = -102-4·4.905·(-30)
Δ = 688.6
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{-(-10)-\sqrt{688.6}}{2*4.905}=\frac{10-\sqrt{688.6}}{9.81}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+\sqrt{688.6}}{2*4.905}=\frac{10+\sqrt{688.6}}{9.81}
$