300=16t^2+20t+300

Solution for 300=16t^2+20t+300 equation:

300=16t^2+20t+300

We move all terms to the left:

300-(16t^2+20t+300)=0

We get rid of parentheses

-16t^2-20t-300+300=0

We add all the numbers together, and all the variables

-16t^2-20t=0

a = -16; b = -20; c = 0;
Δ = b2-4ac
Δ = -202-4·(-16)·0
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-20}{2*-16}=\frac{0}{-32}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+20}{2*-16}=\frac{40}{-32}
=-1+1/4
$