0=300-4.9(t)^2

Solution for 0=300-4.9(t)^2 equation:

0=300-4.9(t)^2

We move all terms to the left:

0-(300-4.9(t)^2)=0

We add all the numbers together, and all the variables

-(300-4.9t^2)=0

We get rid of parentheses

4.9t^2-300=0

a = 4.9; b = 0; c = -300;
Δ = b2-4ac
Δ = 02-4·4.9·(-300)
Δ = 5880
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:

$\sqrt{\Delta}=\sqrt{5880}=\sqrt{196*30}=\sqrt{196}*\sqrt{30}=14\sqrt{30}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-14\sqrt{30}}{2*4.9}=\frac{0-14\sqrt{30}}{9.8}
=-\frac{14\sqrt{30}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+14\sqrt{30}}{2*4.9}=\frac{0+14\sqrt{30}}{9.8}
=\frac{14\sqrt{30}}{9.8}
$