-4.905t^2+18t+12=0

Solution for -4.905t^2+18t+12=0 equation:

-4.905t^2+18t+12=0

a = -4.905; b = 18; c = +12;
Δ = b2-4ac
Δ = 182-4·(-4.905)·12
Δ = 559.44
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{-(18)-\sqrt{559.44}}{2*-4.905}=\frac{-18-\sqrt{559.44}}{-9.81}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+\sqrt{559.44}}{2*-4.905}=\frac{-18+\sqrt{559.44}}{-9.81}
$