GIVEN: $\sqrt{3-2x} = \sqrt{2x} +1$

Square both sides to get: $(\sqrt{3-2x})^2 =(\sqrt{2x} +1)^2$

In other words: $(\sqrt{3-2x})^2 =(\sqrt{2x} +1)(\sqrt{2x} +1)$

Use FOIL to expand right side: $(\sqrt{3-2x})^2 =(\sqrt{2x})^2 + 1\sqrt{2x} + 1\sqrt{2x} + 1^2$

Simplify right side to get: $3-2x=2x+2\sqrt{2x}+1$

Subtract 1 from both sides to get: $2-2x=2x+2\sqrt{2x}$

Subtract 2x from both sides to get: $2-4x=2\sqrt{2x}$

Divide both sides by 2 to get: $1-2x=\sqrt{2x}$

Square both sides to get: $(1-2x)^2=(\sqrt{2x})^2$

Expand and simplify to get: $1-4x+4x^2=2x$

Add 4x to both sides: $1 + 4x^2=6x$

Subtract 1 from both sides to get: $4x^2=6x-1$

Answer: E

Cheers,
Brent