-4.9t^2+12t+6=0

Solution for -4.9t^2+12t+6=0 equation:

-4.9t^2+12t+6=0

a = -4.9; b = 12; c = +6;
Δ = b2-4ac
Δ = 122-4·(-4.9)·6
Δ = 261.6
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{-(12)-\sqrt{261.6}}{2*-4.9}=\frac{-12-\sqrt{261.6}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+\sqrt{261.6}}{2*-4.9}=\frac{-12+\sqrt{261.6}}{-9.8}
$