0=-4.9t^2+12t+3

Solution for 0=-4.9t^2+12t+3 equation:

0=-4.9t^2+12t+3

We move all terms to the left:

0-(-4.9t^2+12t+3)=0

We add all the numbers together, and all the variables

-(-4.9t^2+12t+3)=0

We get rid of parentheses

4.9t^2-12t-3=0

a = 4.9; b = -12; c = -3;
Δ = b2-4ac
Δ = -122-4·4.9·(-3)
Δ = 202.8
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{-(-12)-\sqrt{202.8}}{2*4.9}=\frac{12-\sqrt{202.8}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+\sqrt{202.8}}{2*4.9}=\frac{12+\sqrt{202.8}}{9.8}
$