4.905t^2-35t+30=0

Solution for 4.905t^2-35t+30=0 equation:

4.905t^2-35t+30=0

a = 4.905; b = -35; c = +30;
Δ = b2-4ac
Δ = -352-4·4.905·30
Δ = 636.4
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{-(-35)-\sqrt{636.4}}{2*4.905}=\frac{35-\sqrt{636.4}}{9.81}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-35)+\sqrt{636.4}}{2*4.905}=\frac{35+\sqrt{636.4}}{9.81}
$