Solution of an inequality does not match with the graph of the function I want to solve the inequality $2(x- \sqrt{2x^{2}-3x-5})\geq 1$.
First I find where the root square is well defined, that is, where $2x^{2}-3x-5 = (x+1)(2x-5) \geq 0.$
Doing a table of sings I get $\sqrt{2x^{2}-3x-5}$ is well defined in $(-\infty, -1] \cup [\frac{5}{2}, \infty)$.
Now, solving the inequality $2(x- \sqrt{2x^{2}-3x-5})\geq 1$, I do the following
$2x-2\sqrt{2x^{2}-3x-5}\geq 1$
$2x-1 \geq 2\sqrt{2x^{2}-3x-5}$
then
$(2x-1)^{2} \geq 4(2x^{2}-3x-5)$
then 
$0 \geq 4x^{2}-8x-21 = (2x-7)(2x+3)$
doing  a table of sings I get that the solutions for this inequality is $[-\frac{3}{2},\frac{7}{2}]$.
Finally, intersecting with the set where $\sqrt{2x^{2}-3x-5}$ is well defined I get the solution set is $[-\frac{3}{2},-1] \cup [\frac{5}{2},\frac{7}{2}]$.
But when I plot the function $2x-2\sqrt{2x^{2}-3x-5}- 1$ I get that the solution is $[\frac{5}{2},\frac{7}{2}]$.

What am I doing wrong? 

When you have
$$2x-1\ge 2\sqrt{2x^2-3x-5}\ \ (\ge 0)$$
you have to have
$$2x-1\ge 0.$$
