12=0t+0.5+9.8t^2

Solution for 12=0t+0.5+9.8t^2 equation:

12=0t+0.5+9.8t^2

We move all terms to the left:

12-(0t+0.5+9.8t^2)=0

We get rid of parentheses

-9.8t^2-0t-0.5+12=0

We add all the numbers together, and all the variables

-9.8t^2-1t+11.5=0

a = -9.8; b = -1; c = +11.5;
Δ = b2-4ac
Δ = -12-4·(-9.8)·11.5
Δ = 451.8
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{-(-1)-\sqrt{451.8}}{2*-9.8}=\frac{1-\sqrt{451.8}}{-19.6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{451.8}}{2*-9.8}=\frac{1+\sqrt{451.8}}{-19.6}
$