12=0+4.9(t)^2

Solution for 12=0+4.9(t)^2 equation:

12=0+4.9(t)^2

We move all terms to the left:

12-(0+4.9(t)^2)=0

We get rid of parentheses

-4.9t^2-0+12=0

We add all the numbers together, and all the variables

-4.9t^2+12=0

a = -4.9; b = 0; c = +12;
Δ = b2-4ac
Δ = 02-4·(-4.9)·12
Δ = 235.2
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{-(0)-\sqrt{235.2}}{2*-4.9}=\frac{0-\sqrt{235.2}}{-9.8}
=-\frac{\sqrt{}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{235.2}}{2*-4.9}=\frac{0+\sqrt{235.2}}{-9.8}
=\frac{\sqrt{}}{-9.8}
$