Question
$$\frac{9}{8}\cdot(4-\frac{1}{2a}) = 3(\frac{1}{4}a+\frac{5}{2})$$
Answer
$$ \begin{matrix}a_1 = 0 & a_2 = \dfrac{ 7 }{ 48 }-\dfrac{\sqrt{ 3889 }}{ 48 } & a_3 = \dfrac{ 7 }{ 48 }+\dfrac{\sqrt{ 3889 }}{ 48 } \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{9}{8}\cdot(4-\frac{1}{2a}) &= 3(\frac{1}{4}a+\frac{5}{2})&& \text{simplify left and right hand side} \\[1 em]\frac{9}{8}\frac{8a-1}{2a} &= 3(\frac{a}{4}+\frac{5}{2})&& \\[1 em]\frac{72a-9}{16a} &= 3 \cdot \frac{a+10}{4}&& \\[1 em]\frac{72a-9}{16a} &= \frac{3a+30}{4}&& \text{multiply ALL terms by } \color{blue}{ 16a\cdot4 }. \\[1 em]16a\cdot4 \cdot \frac{72a-9}{16a} &= 16a\cdot4 \cdot \frac{3a+30}{4}&& \text{cancel out the denominators} \\[1 em]288a^3-36a^2 &= 48a^2+480a&& \text{move all terms to the left hand side } \\[1 em]288a^3-36a^2-48a^2-480a &= 0&& \text{simplify left side} \\[1 em]288a^3-84a^2-480a &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ 288x^{3}-84x^{2}-480x = 0 } $, first we need to factor our $ x $.
$$ 288x^{3}-84x^{2}-480x = x \left( 288x^{2}-84x-480 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The remaining roots can be found by solving equation $ 288x^{2}-84x-480 = 0$.
$ 288x^{2}-84x-480 = 0 $ is a quadratic equation.
You can use step-by-step quadratic equation solver to see a detailed explanation on how to solve this equation.
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