Show that there are two values of time for which a projectile is at the same . . .

Question : Show that there are two values of time for which a projectile is at the same height. Also show that the sum of these two times is equal to the time of flight.

Doubt by Yana 

Solution : 

Consider the vertical motion of the projectile 
s=ut+½at²
s_y=u_yt+½a_yt²
y=usinθt+½(-g)t²
y=usinθt-½gt²
½gt²-usinθt+y=0
Multiply both sides by 2
gt²-2usinθt+2y=0

This is a quadratic equation with variable t.

Here 
a=g
b=-2usinθ

c=2y

D(Discriminate) = b²-4ac
D=(-2usinθ)²-4(g)(2y)
D=4u²sin²θ-8gy
D>0

So, except at the peak and initial/final points, there are two different times for the same height h one during ascent (going up) and one during descent (going down).

Let these two times be t₁ and t₂

We know 
sum of zeroes = -b/a
t₁+t₂ = -(-2usinθ)/g
t₁+t₂=2using/g
t₁+t₂=T (Time of flight) 

Hence Proved.