0=368-16t^2

Solution for 0=368-16t^2 equation:

0=368-16t^2

We move all terms to the left:

0-(368-16t^2)=0

We add all the numbers together, and all the variables

-(368-16t^2)=0

We get rid of parentheses

16t^2-368=0

a = 16; b = 0; c = -368;
Δ = b2-4ac
Δ = 02-4·16·(-368)
Δ = 23552
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:

$\sqrt{\Delta}=\sqrt{23552}=\sqrt{1024*23}=\sqrt{1024}*\sqrt{23}=32\sqrt{23}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-32\sqrt{23}}{2*16}=\frac{0-32\sqrt{23}}{32}
=-\frac{32\sqrt{23}}{32}
=-\sqrt{23}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+32\sqrt{23}}{2*16}=\frac{0+32\sqrt{23}}{32}
=\frac{32\sqrt{23}}{32}
=\sqrt{23}
$