Math help

Assuming parabolic motion, the horizontal velocity vector of a projectile is v(0) cos(θ. The vertical velocity vector is modeled by v(0) sin(θ t + gt. Therefore, we integrate to find the positions:

∫ v(t) dt = ∫ v(0) cos(θ dt = x(t)
v(0) cos(θ t +x(0) = x(t)

∫ v(t) dt = ∫ v(0) sin(θ +gt dt = y(t)
v(0) sin(θ t +gt^2/2 +r(0) = y(t)

Since x(0) is 0 at my starting point, I'm going to rewrite this as:
v(0) cos(θ t = x(t)

I'm trying to find a general solution that allows for me to clear a wall of height h distance d away. Thus, x(t) = d and y(t) = h. Therefore, my problem is to eliminate t. If I rearrange x(t), I get:

t = d /(v(0) cos(θ)

Subbing into y(t), I get:

g(d /(v(0) cos(θ))^2/2 +v(0) sin(θ (d /(v(0) cos(θ)) +r(0) = h

g(d /(v(0) cos(θ))^2/2 +v(0) sin(θ (d /(v(0) cos(θ)) +r(0) -h = 0

Because I have to clear the wall:

g (d /(v(0) cos(θ) )^2/2 +v(0) sin(θ (d /(v(0) cos(θ) ) +r(0) -h > 0

Simplifying:

gd^2 (sec(θ)^2 /(2 v(0)^2) + d tan(θ + r(0) -h > 0

gd^2 (1+(tan(θ)^2) /(2 v(0)^2) + d tan(θ + r(0) -h > 0

gd^2 (tan(θ)^2 /(2 v(0)^2) + d tan(θ + r(0) -h + gd^2 /(2v(0)^2) > 0

Now I can use the quadratic formula to solve for tan(θ.

tan(θ > ( -d +/- √(d^2 -4(gd^2/(2v(0)^2)(gd^2 /(2v(0)^2) +r(0) -h)) ) /(gd^2 /v(0)^2)

tan(θ > ( -d +/- √(d^2(1 -(2g/v(0)^2)(gd^2 /(2v(0)^2) +r(0) -h)) ) /(gd^2 /v(0)^2)

tan(θ > ( -d +/- d√((1 -(2g/v(0)^2)(gd^2 /(2v(0)^2) +r(0) -h)) ) /(gd^2 /v(0)^2)

tan(θ > (v(0)^2) ( -1 +/- √((1 -(2g/v(0)^2)(gd^2 /(2v(0)^2) +r(0) -h)) ) /(gd)

tan(θ > (-v(0)^2 +/- v(0)√((v(0)^2 -(2g)(gd^2 /(2v(0)^2) +r(0) -h)) ) /(gd)

tan(θ > (-v(0)^2 +/- v(0)√((v(0)^2 -(gd/v(0))^2 -(2g)(r(0) -h)) ) /(gd)

θ > arctan((-v(0)^2 +/- v(0)√((v(0)^2 -(gd/v(0))^2 -(2g)(r(0) -h)) ) /(gd)

When I put this into my calculator, however, it is giving me incorrect solutions. I don't know what I did wrong.