Answer
$$x=6$$
Work Step by Step
$$\sqrt{2x-3}=x-3$$
Square both sides: $$(\sqrt{2x-3})^{2} = (x-3)^{2}$$ $$2x-3 = x^{2} - 6x + 9$$
Simplify: $$x^{2} - 8x + 12 = 0$$
Use the quadratic formula to solve for $x$.
$a=1$, $b=-8$,$c=12$ $$x = \frac{-b±\sqrt{b^2-4ac}}{2a}$$ $$x = \frac{-(-8)±\sqrt{(-8)^2-4(1)(12)}}{2(1)}$$ $$x =6$$ $$x =2$$
Verify the solution by substituting the values of $x$ to the original equation:
if $x=6$ $$\sqrt{2(6)-3}=6-3$$ $$\sqrt{12}=3$$ $$3=3$$--> TRUE
if $x = 2$ $$\sqrt{2(2)-3}=2-3$$ $$\sqrt{4-3} = -1$$ $$\sqrt1=-1$$ $$1=-1$$--> FALSE
Hence, $x=6$.
