0=-16t^2+4t+442

Solution for 0=-16t^2+4t+442 equation:

0=-16t^2+4t+442

We move all terms to the left:

0-(-16t^2+4t+442)=0

We add all the numbers together, and all the variables

-(-16t^2+4t+442)=0

We get rid of parentheses

16t^2-4t-442=0

a = 16; b = -4; c = -442;
Δ = b2-4ac
Δ = -42-4·16·(-442)
Δ = 28304
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:

$\sqrt{\Delta}=\sqrt{28304}=\sqrt{16*1769}=\sqrt{16}*\sqrt{1769}=4\sqrt{1769}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{1769}}{2*16}=\frac{4-4\sqrt{1769}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{1769}}{2*16}=\frac{4+4\sqrt{1769}}{32}
$