Indefinite integral of absolute value given x(0)

Question

*edited- the mistake was I took $\sqrt{2x-3}$ and not $\sqrt{3-2x}$. When you take the correct one you get: $$(t-c)^2 = 3 - 2x$$ $$ c=\sqrt3$$ $$=> x = \sqrt3t-{t^2\over2}$$ $$---------------------------------------$$ I have this equation: $$x'(t)=\sqrt{|2x-3|}$$ and $$x(0)=0$$ I have to find the semi-regular solution of this ODE. Because $x(0)=0$ I know that $x < 1.5$ for all $t$.So my integral is : $$t = \int{{dx\over\sqrt{2x-3}}}$$ $$=> t = (2x-3)^{1\over2}+c$$ And here I am a bit stuck because I get $$({t} - {c})^2 = 2x-3$$ $$({t} - {c})^2 + 3 = 2x$$but $x(0)=0$ and the left side is always greater than $0$... How do I proceed from here?