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This is a part of a kinetic equation of motion,

\begin{align} x(t) - x_0 &= \frac{1}{2}a \left[ \frac{v(t) - v_0}{a} \right]^2 + v_0 \left[ \frac{v(t) - v_0}{a} \right] \tag{1.17} \\ &= \frac{v(t)^2 - v_0^2}{2a} \, . \tag{1.18} \end{align}

I am interested in how one passes from 1.17 to 1.18. I am not sure if this is the proper format for a question.

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    $\begingroup$ Welcome to Physics Stack Exchange. Please note that this site supports mathjax, a nice way to show mathematical notation in your question and answers. We much prefer to have equations written with mathjax rather than using a picture. I edited this question to use mathjax, so now if you hit the "edit" button you can see how it works. $\endgroup$
    – DanielSank
    Oct 11, 2015 at 3:32

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Factor the first term: $$\Delta x= \dfrac{v(t)-v_0}{2}\left[\dfrac{v(t)-v_0}{a}\right] + v_0\left[\dfrac{v(t)-v_0}{a}\right]$$ $$= \left(\dfrac{v(t)-v_0+2v_0}{2}\right)\left[\dfrac{v(t)-v_0}{a}\right]$$ $$= \left(\dfrac{v(t)+ v_0}{2}\right)\left[\dfrac{v(t)-v_0}{a}\right]$$ $$\Delta x= \dfrac{v(t)^2-v_0^2}{2a}$$

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