Solve the inequality $ 2x^{2} + x - 4 \sqrt{2x^{2} + x + 4} < 1 $ 
Solve the inequality
  $$ 2x^{2} + x - 4 \sqrt{2x^{2} + x + 4} < 1 $$


Attempt:
I get that $2x^{2} + x + 4 > 0$ for all $x$. Let $y = 2x^{2} + x$, then the inequality becomes
$$ y - 4 \sqrt{y + 4} < 1 $$
$$ 0 < \frac{1 + 4 \sqrt{y+4} - y}{y}$$
so from here I get that $0 < y < 9 - 2\sqrt{14} \: \:\cup \:\: y > 9 + 2\sqrt{14}$. This is for $y>0$ or $x(2x+1)>0 \implies x > 0 \:\: \cup \:\: x < -1/2$
I may continue to solve for $x$.. but the answer key says that the solution is $ -7/2 < x < 3$. I doubt that the solution set will be that tidy. Are there better approaches to solve the inequality?

It's $$2x^2+x+4-4\sqrt{2x^2+x+4}+4<9$$ or
$$\left(\sqrt{2x^2+x+4}-2\right)^2<9$$ or since $\sqrt{2x^2+x+4}>0$, it's
$$\sqrt{2x^2+x+4}<5$$ or
$$2x^2+x-21<0$$ or
$$(x-3)(2x+7)<0$$ or
$$-3.5<x<3.$$
