I'm having a hard time to find the radius of this sphere equation. I got the center correct, but I can't get the correct answer for the radius. I'm completing the square, but my solution is off. I guess I have some algebra error somewhere along the way.
$$(2x^2-2x)+(2y^2-3y)+(2z^2+5z)=2$$ For $x$:
$$2x^2-2x=0$$ $$(x-\frac{1}{2})^2=1/4$$
For $y$: $$2y^2-3y=0$$ $$(y-\frac{3}{4})^2=9/16$$
For $z$:
$$2z^2+5z=0$$ $$(z+\frac{5}{4})^2=25/16$$
Which gives me
$$(x-\frac{1}{2})^2+(y-\frac{3}{4})^2+(z+\frac{5}{4})^2=2+\frac{1}{4}+\frac{9}{16}+\frac{25}{16}$$
$$(x-\frac{1}{2})^2+(y-\frac{3}{4})^2+(z+\frac{5}{4})^2=\frac{70}{16}$$
But the correct answer is $\frac{54}{16}$. Which then becomes $r=3\frac{\sqrt6}{4}$
Where is my mistake?
Your commenters have it precisely correct. You need to divide across completely by two before completing the square. After dividing across both sides of the equality by two you should be staring at:
$$(x^2-x)+(y^2-\frac{3}{2}y)+(z^2+\frac{5}{2}z)=1.$$
Now when you complete your squares (using the same method you are currently using) you will get the correct constant and radius.
Your actual error was in thinking you could pull your pieces off, set them equal to zero and complete the square on them individually. Notice how the way you are doing this you are dividing each term by 2 to get a 1 coefficient on the square term, and assuming the right hand side is $\frac{0}{2}=0$. It is not sufficient. You actually need to divide both sides of the entire original equality by two before proceeding with your scheme. Keep staring, you will see what we mean.