<s>[INST] 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?
 [/INST] Let $y= \sqrt{2x^{2} + x + 4} \geq 0$, (notice that $2x^2+x+4>0$ for all real $x$) then we get $$
y^2-4-4y<1\implies y^2-4y-5 = (y-5)(y+1)<0$$
So $$y\in (-1,5)$$
and so 
$$ \sqrt{2x^{2} + x + 4} < 5 \implies 2x^2+x+4<25$$
so $$2x^2+x-21= (2x+7)(x-3)<0 \implies x\in (-{7\over 2},3)$$
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