<s>[INST] Solve this equation: $(x+2)(\sqrt{2x+3}-2\sqrt{x+1})+\sqrt{2x^2+5x+3}=1$ Solve this equation: $(x+2)(\sqrt{2x+3}-2\sqrt{x+1})+\sqrt{2x^2+5x+3}=1$

This is my try

Let $t=\sqrt{2x+3}-2\sqrt{x+1}$ or $t^2=6x+7-4\sqrt{2x^2+5x+3}=1$
The equation is equivalent to: $t^2-4(x+2)t-6x-3=0\qquad(*)$

I have no idea how to solve the equation $(*)$. Who can help me?
 [/INST] $$(x+2)(\sqrt{2x+3}-2\sqrt{x+1})+\sqrt{2x^2+5x+3}=1$$
$$(x+2)(\sqrt{2x+3}-2\sqrt{x+1})=1-\sqrt{2x^2+5x+3}$$
$$(x+2)^2(6x+7-4\sqrt{(2x+3)(x+1)})=2x^2+5x+4-2\sqrt{2x^2+5x+3}$$
$$(x+2)^2(6x+7)-(2x^2+5x+4)=2(2(x+2)^2-1)\sqrt{(2x+3)(x+1)})$$
$$(6 x^3+29 x^2+47 x+24)^2=4(2x^2+8x+7)^2(2x+3)(x+1)$$
Latter can be factored as roots $-1,\,-\frac{1}{2},3$ are known.

$\frac{4 x^6+12 x^5-19 x^4-106 x^3-135 x^2-68 x-12 }{(x+\frac{1}{2})(x+1)(x-3)}$$=2 (2 x+1) (x+2)^2$
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