0=300+10t-4.9t^2

Solution for 0=300+10t-4.9t^2 equation:

0=300+10t-4.9t^2

We move all terms to the left:

0-(300+10t-4.9t^2)=0

We add all the numbers together, and all the variables

-(300+10t-4.9t^2)=0

We get rid of parentheses

4.9t^2-10t-300=0

a = 4.9; b = -10; c = -300;
Δ = b2-4ac
Δ = -102-4·4.9·(-300)
Δ = 5980
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:

$\sqrt{\Delta}=\sqrt{5980}=\sqrt{4*1495}=\sqrt{4}*\sqrt{1495}=2\sqrt{1495}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{1495}}{2*4.9}=\frac{10-2\sqrt{1495}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{1495}}{2*4.9}=\frac{10+2\sqrt{1495}}{9.8}
$