4.905t^2+3t-10=0

Solution for 4.905t^2+3t-10=0 equation:

4.905t^2+3t-10=0

a = 4.905; b = 3; c = -10;
Δ = b2-4ac
Δ = 32-4·4.905·(-10)
Δ = 205.2
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{-(3)-\sqrt{205.2}}{2*4.905}=\frac{-3-\sqrt{205.2}}{9.81}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{205.2}}{2*4.905}=\frac{-3+\sqrt{205.2}}{9.81}
$