12t+4.9t^2-1=0

Solution for 12t+4.9t^2-1=0 equation:

12t+4.9t^2-1=0

a = 4.9; b = 12; c = -1;
Δ = b2-4ac
Δ = 122-4·4.9·(-1)
Δ = 163.6
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{163.6}}{2*4.9}=\frac{-12-\sqrt{163.6}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+\sqrt{163.6}}{2*4.9}=\frac{-12+\sqrt{163.6}}{9.8}
$