0=-16t^2+352t+368

Solution for 0=-16t^2+352t+368 equation:

0=-16t^2+352t+368

We move all terms to the left:

0-(-16t^2+352t+368)=0

We add all the numbers together, and all the variables

-(-16t^2+352t+368)=0

We get rid of parentheses

16t^2-352t-368=0

a = 16; b = -352; c = -368;
Δ = b2-4ac
Δ = -3522-4·16·(-368)
Δ = 147456
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{147456}=384$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-352)-384}{2*16}=\frac{-32}{32}
=-1
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-352)+384}{2*16}=\frac{736}{32}
=23
$