0=4.9t^2+8t-3

Solution for 0=4.9t^2+8t-3 equation:

0=4.9t^2+8t-3

We move all terms to the left:

0-(4.9t^2+8t-3)=0

We add all the numbers together, and all the variables

-(4.9t^2+8t-3)=0

We get rid of parentheses

-4.9t^2-8t+3=0

a = -4.9; b = -8; c = +3;
Δ = b2-4ac
Δ = -82-4·(-4.9)·3
Δ = 122.8
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{-(-8)-\sqrt{122.8}}{2*-4.9}=\frac{8-\sqrt{122.8}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+\sqrt{122.8}}{2*-4.9}=\frac{8+\sqrt{122.8}}{-9.8}
$