Solve the equation $18x^2-18x \sqrt{x}-17x-8 \sqrt{x}-2=0$ 
Solve the following equation. $$18x^2-18x \sqrt{x}-17x-8 \sqrt{x}-2=0.$$ 

Taking $\sqrt{x}=t$ we get equivalent equation $18t^4 -18t^3 - 17t^2-8t-2=0$.
From this point I have tried to factor it , write RHS as sum of two squares and its variants but nothing seem to work. Then putting the original equation in wolfram alpha I got solution $x=\frac{2}{9}(7+2 \sqrt{10})$. Can anyone suggest a method to solve it without wolfram alpha or any such computer method. Thanks in advance.

First, the condition is $x\geq 0$.
Now, as you did, put $t = \sqrt{x}$, then we have the equation.
$$18t^4-18t^3 -17t^2-8t-2 = 0.$$
Obviously, $t=0$ is not the solution of this equation. Then, we can divide two sides by $t^4$, $$18-\frac{18}{t}-\frac{17}{t^2}-\frac{8}{t^3}-\frac{2}{t^4} = 0$$
Let $u = \frac{1}{t}$, we have
$$18-18u-17u^2-8u^3-2u^4=0$$
$$25 - 5(u+1)^2 - 2(u+1)^4 = 0$$
Now, you can solve the quadratic equation of $(u+1)^2$:
$$(u+1)^2 = \frac{5}{2}$$
Thus, $u = \frac{\sqrt{5}}{\sqrt{2}}-1 = \frac{\sqrt{10}-2}{2}$.
So, $t = \frac{1}{u} = \frac{\sqrt{10}+2}{3}$. Finally, $x = t^2 = \frac{14+4\sqrt{10}}{9}.$
