4.905t^2+10t-45=0

Solution for 4.905t^2+10t-45=0 equation:

4.905t^2+10t-45=0

a = 4.905; b = 10; c = -45;
Δ = b2-4ac
Δ = 102-4·4.905·(-45)
Δ = 982.9
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{982.9}}{2*4.905}=\frac{-10-\sqrt{982.9}}{9.81}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+\sqrt{982.9}}{2*4.905}=\frac{-10+\sqrt{982.9}}{9.81}
$