$$5c+\frac{38}{4}c+40-1 = \frac{1}{c}+4$$
Answer
$$ \begin{matrix}c_1 = -\dfrac{ 35 }{ 29 }-\dfrac{\sqrt{ 1283 }}{ 29 } & c_2 = -\dfrac{ 35 }{ 29 }+\dfrac{\sqrt{ 1283 }}{ 29 } \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} 5c+\frac{38}{4}c+40-1 &= \frac{1}{c}+4&& \text{multiply ALL terms by } \color{blue}{ 4c }. \\[1 em]4c\cdot5c+4c\frac{38}{4}c+4c\cdot40-4c\cdot1 &= 4c\cdot\frac{1}{c}+4c\cdot4&& \text{cancel out the denominators} \\[1 em]20c^2+38c^2+160c-4c &= 4+16c&& \text{simplify left and right hand side} \\[1 em]58c^2+156c &= 16c+4&& \text{move all terms to the left hand side } \\[1 em]58c^2+156c-16c-4 &= 0&& \text{simplify left side} \\[1 em]58c^2+140c-4 &= 0&& \\[1 em] \end{aligned} $$
$ 58x^{2}+140x-4 = 0 $ is a quadratic equation.
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