Question
$$130^2 = 90\cdot(1+\frac{500^2}{c^2})$$
Answer
$$ c = 0 $$
Explanation
$$ \begin{aligned} 130^2 &= 90\cdot(1+\frac{500^2}{c^2})&& \text{simplify left and right hand side} \\[1 em]16900 &= 90\cdot(1+\frac{250000}{c^2})&& \\[1 em]16900 &= 90 \cdot \frac{c^2+250000}{c^2}&& \\[1 em]16900 &= \frac{90c^2+22500000}{c^2}&& \text{multiply ALL terms by } \color{blue}{ c^2 }. \\[1 em]c^2\cdot16900 &= c^2\frac{90c^2+22500000}{c^2}&& \text{cancel out the denominators} \\[1 em]16900c^2 &= 90c^6+22500000c^4&& \text{move all terms to the left hand side } \\[1 em]16900c^2-90c^6-c^4 &= 0&& \text{simplify left side} \\[1 em]-90c^6-c^4+16900c^2 &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ -90x^{6}-x^{4}+16900x^{2} = 0 } $, first we need to factor our $ x^2 $.
$$ -90x^{6}-x^{4}+16900x^{2} = x^2 \left( -90x^{4}-x^{2}+16900 \right) $$$ x = 0 $ is a root of multiplicity $ 2 $.
The remaining roots can be found by solving equation $ -90x^{4}-x^{2}+16900 = 0$.
This polynomial has no rational roots that can be found using Rational Root Test.
Roots were found using quartic formulas
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