Isolating x in a quadratic expression by completing the square

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I came across some study notes regarding quadratic expressions and there is a solution that I am having a hard time understanding. In the study notes it is stated: "The variable x in the quadratic expression $ax^2 + bx$, where a is nonzero, may be isolated by the technique of completing the square". Then there is the following example give:

$$\displaystyle2x^2 + 3x = 2\left(x^2 + \frac{3x}{2}\right) = 2\left(x^2 + \frac{3x}{2} + \frac{9}{16} - \frac{9}{16}\right) = 2\left(x + \frac{3}{4}\right)^2 - \frac{9}{8}$$

What I am having a hard time understanding is in $2\left(x^2 + \frac{3x}{2}\right) = 2\left(x^2 + \frac{3x}{2} + \frac{9}{16} - \frac{9}{16}\right)$ where are the $+ 9/16$ and $- 9/16$ values coming from?

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As noticed in the comments, the $\frac 9{16}$ is found by inspection from the $\frac 3{2}x$ term in such way that

$$2 \cdot \sqrt{\frac 9{16}}\cdot x= \frac 3{2}x$$

indeed

$$(x+c)^2=x^2+2cx+c^2$$

Therefore, in general, to complete the square starting from $ax^2+bx$ we can proceed as follows

$$ax^2+ bx=a\left(x^2+\frac b ax\right)=a\left(x^2+\frac b ax+c^2\right)-ac^2=a\left[(x+ c)^2-c^2\right]$$

such that

$$2c=\frac b a \implies c=\frac{b}{2a}\quad c^2=\frac{b^2}{4a^2}$$