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√(2x+1)+√(x-3)=2√(x)

(2x+1)+2√(2x+1)√(x-3)+(x-3)=4x

2√(2x+1)√(x-3)=x+2

4(2x+1)(x-3)=(x+2)^2

(±)^2(8x^2-20x-12)=(±)^2(x^2+4x+4)

(±)^2(7x^2-24x-16)=0

(±)^2(x^2-(24/7)x)=(±^2)(16/7)

(±)^2(x^2-(24/7)x+(144/7^2))=(±)^2((144/7^2)+(16/7))

[(±)^4][x-(12/7)]^2=(±)^2[(144+112)/7^2]

±√([(±)^4][(x-(12/7)]^2)=±√{(±)^2[256/7^2]}

±±(±)^2(x-(12/7))=±±16/7

x-(12/7)=16/7

x=(12+16)/7

x=28/7

x=4

$\begin{aligned} \sqrt{2x+1} + \sqrt{x-3} & = 2\sqrt{x} \quad \quad \small \color{#3D99F6}{\text{Squaring both sides}} \\ 2x+1 + 2\sqrt{(2x+1)(x-3)} + x-3 & = 4x \\ 2\sqrt{(2x+1)(x-3)} & = x +2 \quad \quad \small \color{#3D99F6}{\text{Squaring both sides again}} \\ 4(2x^2-5x-3) & = x^2+4x+4 \\ 7x^2-24x-16 & = 0 \\ (7x+4)(x-4) & = 0 \\ \Rightarrow x & = \boxed{4} \quad \quad \small \color{#3D99F6}{x = - \frac{4}{7} \text{ is rejected because } 2\sqrt{x}} \\ & \quad \quad \quad \quad \small \color{#3D99F6}{\text{ in RHS of the original equation is undefined.}} \end{aligned}$