0=4.9t^2+1.43t-12

Solution for 0=4.9t^2+1.43t-12 equation:

0=4.9t^2+1.43t-12

We move all terms to the left:

0-(4.9t^2+1.43t-12)=0

We add all the numbers together, and all the variables

-(4.9t^2+1.43t-12)=0

We get rid of parentheses

-4.9t^2-1.43t+12=0

a = -4.9; b = -1.43; c = +12;
Δ = b2-4ac
Δ = -1.432-4·(-4.9)·12
Δ = 237.2449
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{-(-1.43)-\sqrt{237.2449}}{2*-4.9}=\frac{1.43-\sqrt{237.2449}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1.43)+\sqrt{237.2449}}{2*-4.9}=\frac{1.43+\sqrt{237.2449}}{-9.8}
$