148=12t+4.905t^2

Solution for 148=12t+4.905t^2 equation:

148=12t+4.905t^2

We move all terms to the left:

148-(12t+4.905t^2)=0

We get rid of parentheses

-4.905t^2-12t+148=0

a = -4.905; b = -12; c = +148;
Δ = b2-4ac
Δ = -122-4·(-4.905)·148
Δ = 3047.76
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{3047.76}}{2*-4.905}=\frac{12-\sqrt{3047.76}}{-9.81}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+\sqrt{3047.76}}{2*-4.905}=\frac{12+\sqrt{3047.76}}{-9.81}
$