0=50+3t-4.9t^2

Solution for 0=50+3t-4.9t^2 equation:

0=50+3t-4.9t^2

We move all terms to the left:

0-(50+3t-4.9t^2)=0

We add all the numbers together, and all the variables

-(50+3t-4.9t^2)=0

We get rid of parentheses

4.9t^2-3t-50=0

a = 4.9; b = -3; c = -50;
Δ = b2-4ac
Δ = -32-4·4.9·(-50)
Δ = 989
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{989}=\sqrt{1*989}=\sqrt{1}*\sqrt{989}=1\sqrt{989}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-1\sqrt{989}}{2*4.9}=\frac{3-1\sqrt{989}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+1\sqrt{989}}{2*4.9}=\frac{3+1\sqrt{989}}{9.8}
$