0=-16t^2+400+2400

Solution for 0=-16t^2+400+2400 equation:

0=-16t^2+400+2400

We move all terms to the left:

0-(-16t^2+400+2400)=0

We add all the numbers together, and all the variables

-(-16t^2+400+2400)=0

We get rid of parentheses

16t^2-400-2400=0

We add all the numbers together, and all the variables

16t^2-2800=0

a = 16; b = 0; c = -2800;
Δ = b2-4ac
Δ = 02-4·16·(-2800)
Δ = 179200
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:

$\sqrt{\Delta}=\sqrt{179200}=\sqrt{25600*7}=\sqrt{25600}*\sqrt{7}=160\sqrt{7}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-160\sqrt{7}}{2*16}=\frac{0-160\sqrt{7}}{32}
=-\frac{160\sqrt{7}}{32}
=-5\sqrt{7}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+160\sqrt{7}}{2*16}=\frac{0+160\sqrt{7}}{32}
=\frac{160\sqrt{7}}{32}
=5\sqrt{7}
$