1900=16t^2+20t+300

Solution for 1900=16t^2+20t+300 equation:

1900=16t^2+20t+300

We move all terms to the left:

1900-(16t^2+20t+300)=0

We get rid of parentheses

-16t^2-20t-300+1900=0

We add all the numbers together, and all the variables

-16t^2-20t+1600=0

a = -16; b = -20; c = +1600;
Δ = b2-4ac
Δ = -202-4·(-16)·1600
Δ = 102800
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:

$\sqrt{\Delta}=\sqrt{102800}=\sqrt{400*257}=\sqrt{400}*\sqrt{257}=20\sqrt{257}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-20\sqrt{257}}{2*-16}=\frac{20-20\sqrt{257}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+20\sqrt{257}}{2*-16}=\frac{20+20\sqrt{257}}{-32}
$