408(t)=-16t^2+408

Solution for 408(t)=-16t^2+408 equation:

408(t)=-16t^2+408

We move all terms to the left:

408(t)-(-16t^2+408)=0

We get rid of parentheses

16t^2+408t-408=0

a = 16; b = 408; c = -408;
Δ = b2-4ac
Δ = 4082-4·16·(-408)
Δ = 192576
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:

$\sqrt{\Delta}=\sqrt{192576}=\sqrt{64*3009}=\sqrt{64}*\sqrt{3009}=8\sqrt{3009}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(408)-8\sqrt{3009}}{2*16}=\frac{-408-8\sqrt{3009}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(408)+8\sqrt{3009}}{2*16}=\frac{-408+8\sqrt{3009}}{32}
$