# To calculate: The solution of the equation 2(\frac{x}{3}+1)^{2}+5(\

To calculate: The solution of the equation $$\displaystyle{2}{\left({\frac{{{x}}}{{{3}}}}+{1}\right)}^{{{2}}}+{5}{\left({\frac{{{x}}}{{{3}}}}+{1}\right)}-{12}={0}.$$

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Calculation:
Consider the equation, $$\displaystyle{2}{\left({\frac{{{x}}}{{{3}}}}+{1}\right)}^{{{2}}}+{5}{\left({\frac{{{x}}}{{{3}}}}+{1}\right)}-{12}={0}.$$
Now, simplify the equation $$\displaystyle{2}{\left({\frac{{{x}}}{{{3}}}}+{1}\right)}^{{{2}}}+{5}{\left({\frac{{{x}}}{{{3}}}}+{1}\right)}-{12}={0}.$$
$$\displaystyle{2}{\left({\frac{{{x}}}{{{3}}}}+{1}\right)}^{{{2}}}+{5}{\left({\frac{{{x}}}{{{3}}}}+{1}\right)}-{12}={0}.$$
$$\displaystyle{2}{\left\lbrace{\left({\frac{{{x}}}{{{3}}}}\right)}^{{{2}}}+{\left({1}\right)}^{{{2}}}+{2}{\left({\frac{{{x}}}{{{3}}}}\right)}{\left({1}\right)}\right\rbrace}+{5}{\left({\frac{{{x}}}{{{3}}}}+{1}\right)}-{12}={0}.$$
$$\displaystyle{2}{\left\lbrace{\left({\frac{{{x}^{{{2}}}}}{{{9}}}}\right)}+{1}+{\frac{{{2}{x}}}{{{3}}}}\right\rbrace}+{\frac{{{5}{x}}}{{{3}}}}+{5}-{12}={0}.$$
$$\displaystyle{\frac{{{2}{x}^{{{2}}}}}{{{9}}}}+{\frac{{{9}{x}}}{{{3}}}}-{5}={0}$$
Now, taking the least common multiple of the denominator,
$$\displaystyle{\frac{{{2}{x}^{{{2}}}}}{{{9}}}}+{\frac{{{9}{x}}}{{{3}}}}-{5}={0}$$
$$\displaystyle{\frac{{{2}{x}^{{{2}}}+{27}{x}-{45}}}{{{9}}}}={0}$$
$$\displaystyle{2}{x}^{{{2}}}+{27}{x}-{45}={0}$$
This is a quadratic equation so compare the equation from the standard form of quadratic equation $$\displaystyle{a}{x}^{{{2}}}+{b}{x}-{c}={0}$$ and identify the values of a, b and c.
Here, a=2, b=27 and c=-45
$$\displaystyle{x}={\frac{{-{\left({27}\right)}\pm\sqrt{{{\left({27}\right)}^{{{2}}}-{4}{\left({2}\right)}{\left(-{45}\right)}}}}}{{{2}{\left({2}\right)}}}}$$
$$\displaystyle={\frac{{-{27}\pm\sqrt{{{729}+{360}}}}}{{{4}}}}$$
$$\displaystyle={\frac{{-{27}\pm\sqrt{{{1089}}}}}{{{4}}}}$$
$$\displaystyle={\frac{{-{27}\pm{33}}}{{{4}}}}$$
Firstly, consider the positive sign,
$$\displaystyle{x}={\frac{{-{27}+{33}}}{{{4}}}}$$
$$\displaystyle={\frac{{{6}}}{{{4}}}}$$
$$\displaystyle={\frac{{{3}}}{{{2}}}}$$ Now, consider the negative sign,
$$\displaystyle{x}={\frac{{-{27}-{33}}}{{{4}}}}$$
$$\displaystyle={\frac{{-{60}}}{{{4}}}}$$
=-15
Now, to check the solution put the values of x in the original equation,
Substitute x=-15 in the equation $$\displaystyle{2}{\left({\frac{{{x}}}{{{3}}}}+{1}\right)}^{{{2}}}+{5}{\left({\frac{{{x}}}{{{3}}}}+{1}\right)}-{12}={0}$$
$$\displaystyle{2}{\left({\frac{{{x}}}{{{3}}}}+{1}\right)}^{{{2}}}+{5}{\left({\frac{{{x}}}{{{3}}}}+{1}\right)}-{12}={0}$$
$$\displaystyle{2}{\left({\frac{{-{15}}}{{{3}}}}+{1}\right)}^{{{2}}}+{5}{\left({\frac{{-{15}}}{{{3}}}}+{1}\right)}-{12}{\overset{{?}}{{=}}}{0}$$
$$\displaystyle{2}{\left(-{5}+{1}\right)}^{{{2}}}+{5}{\left(-{5}+{1}\right)}-{12}{\overset{{?}}{{=}}}{0}$$
$$\displaystyle{2}{\left(-{4}\right)}^{{{2}}}+{5}{\left(-{4}\right)}-{12}{\overset{{?}}{{=}}}{0}$$
Further simplified,
$$\displaystyle{2}{\left({16}\right)}-{20}-{12}{\overset{{?}}{{=}}}{0}$$
$$\displaystyle{32}-{32}{\overset{{?}}{{=}}}{0}$$
0=0
This solution is correct.
Now, substitute $$\displaystyle{x}={\frac{{{3}}}{{{2}}}}$$ in the equation $$\displaystyle{2}{\left({\frac{{{x}}}{{{3}}}}+{1}\right)}^{{{2}}}+{5}{\left({\frac{{{x}}}{{{3}}}}+{1}\right)}-{12}={0}$$
$$\displaystyle{2}{\left({\frac{{{x}}}{{{3}}}}+{1}\right)}^{{{2}}}+{5}{\left({\frac{{{x}}}{{{3}}}}+{1}\right)}-{12}={0}$$
$$\displaystyle{2}{\left({\frac{{{\frac{{{3}}}{{{2}}}}}}{{{3}}}}+{1}\right)}^{{{2}}}+{5}{\left({\frac{{{\frac{{{3}}}{{{2}}}}}}{{{3}}}}+{1}\right)}-{12}{\overset{{?}}{{=}}}{0}$$
$$\displaystyle{2}{\left({\frac{{{1}}}{{{2}}}}+{1}\right)}^{{{2}}}+{5}{\left({\frac{{{1}}}{{{2}}}}+{1}\right)}-{12}{\overset{{?}}{{=}}}{0}$$
$$\displaystyle{2}{\left({\frac{{{3}}}{{{2}}}}\right)}^{{{2}}}+{5}{\left({\frac{{{3}}}{{{2}}}}\right)}-{12}{\overset{{?}}{{=}}}{0}$$
Further simplified,
$$\displaystyle{2}{\left({\frac{{{9}}}{{{4}}}}\right)}+{5}{\left({\frac{{{3}}}{{{2}}}}\right)}-{12}{\overset{{?}}{{=}}}{0}$$
$$\displaystyle{\frac{{{9}}}{{{2}}}}+{\frac{{{15}}}{{{2}}}}-{12}{\overset{{?}}{{=}}}{0}$$
$$\displaystyle{\frac{{{24}-{24}}}{{{2}}}}{\overset{{?}}{{=}}}{0}$$
0=0
This solution is correct.
Therefore, the solution set of the equation $$\displaystyle{2}{\left({\frac{{{x}}}{{{3}}}}+{1}\right)}^{{{2}}}+{5}{\left({\frac{{{x}}}{{{3}}}}+{1}\right)}-{12}={0}$$
is $$\displaystyle{\left\lbrace-{15},\ {\frac{{{3}}}{{{2}}}}\right\rbrace}.$$