<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] $$
(x+2)(\sqrt{2x+3}-2\sqrt{x+1})+\sqrt{2x^2+5x+3}=1
$$
Let $t=x+1$,
$$2x+3=2t+1$$
$$(t+1)(\sqrt{2t+1}-2\sqrt{t})+\sqrt{2t^2+t}=1$$
$$(t+1)(\sqrt{2t+1}-2\sqrt{t})=1-\sqrt{2t^2+t}$$
Squaring both sides,
$$(t^2+2t+1)(2t+1+4t-4\sqrt{2t^2+t})=1+2t^2+t-2\sqrt{2t^2+t}$$
$$6t^3+12t^2+6t+t^2+2t+1-\sqrt{2t^2+t}(-4t^2-8t-2)=1+2t^2+t$$
$$t(6t^2+11t+7)=\sqrt{t}\sqrt{2t+1}(-4t^2-8t-2)$$
Again Squaring both sides,
$$
4t^6-12t^5-19t^4+10t^3+9t^2-4t=0
$$
Therefore , $t=-1,0,\frac{1}{2},4$
$$
x=-2,-1,-\frac{1}{2},3
$$
</s>