$$\frac{5c+38}{4c+40}-1 = \frac{1}{c}+4$$
Answer
$$ \begin{matrix}c_1 = -\dfrac{ 83 }{ 15 }-\dfrac{\sqrt{ 6289 }}{ 15 } & c_2 = -\dfrac{ 83 }{ 15 }+\dfrac{\sqrt{ 6289 }}{ 15 } \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{5c+38}{4c+40}-1 &= \frac{1}{c}+4&& \text{multiply ALL terms by } \color{blue}{ (4c+40)c }. \\[1 em](4c+40)c\frac{5c+38}{4c+40}-(4c+40)c\cdot1 &= (4c+40)c\cdot\frac{1}{c}+(4c+40)c\cdot4&& \text{cancel out the denominators} \\[1 em]5c^2+38c-(4c^2+40c) &= 4c+40+16c^2+160c&& \text{simplify left and right hand side} \\[1 em]5c^2+38c-4c^2-40c &= 16c^2+164c+40&& \\[1 em]c^2-2c &= 16c^2+164c+40&& \text{move all terms to the left hand side } \\[1 em]c^2-2c-16c^2-164c-40 &= 0&& \text{simplify left side} \\[1 em]-15c^2-166c-40 &= 0&& \\[1 em] \end{aligned} $$
$ -15x^{2}-166x-40 = 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.
This page was created using
Equations Solver