<s>[INST] $5x^2−10x+7$ in completed square form? I was studying quadratic equations and practicing to solve them using the technique of completing the squares. 
My answer was as follows: 
$$5x^2−10x=−7$$
$$5x^2−10x+25=18$$
$$x(x-5)^2−5(x-5)=18$$
$$\boldsymbol{(x-5)^2-18=0}$$ 
The answer in the book is: 
$$\boldsymbol{5(x-1)^2+2=0}$$ 
I'm confused. I just started with these sums today. Could you please tell me where I've gone wrong? 
 [/INST] Usually, if it's an equation, then we can divide both sides to get a quadratic term of the form $x^2$ instead of $ax^2$:
\begin{eqnarray*}
5x^2 -10x &=& -7 \\
x^2 - 2x &=& - \frac{7}{5} \\ \\
\left(x-1\right)^2-1 &=& -\frac{7}{5} \\ \\
\left(x-1\right)^2 &=& - \frac{2}{5}
\end{eqnarray*}
This has no real roots. You are welcome - because it's an equation - to multiply both sides by $5$:
$$5(x-1)^2 = -2 \implies 5(x-1)^2 + 2 = 0$$
This is a rather odd mixture of methods. If you're solving the equation $5x^2-10x=-7$ then you can solve from $(x-1)^2 = -\frac{2}{5}$ (allowing complex numbers). There's no need to re-introduce the factor of $5$. If you're completing the square on the expression $5x^2-10x+7$ then you can't just divide by $5$, and you need to take out common factors. Then you get 
$$5x^2-10x+7 \equiv 5(x-1)^2+2$$
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