Question

$2\sqrt{5}{x}^2-3x-\sqrt{5}=0$ Find $x$.

Answer 1

$\begin{array}{c}2\sqrt{5}{x}^2-3x-\sqrt{5}=0\\ \Rightarrow 2\sqrt{5}{x}^2+2x-5x\sqrt{5}=0\\ \Rightarrow 2x\left(\sqrt{5}x+1\right)-\sqrt{5}\left(\sqrt{5}x+1\right)\\ \Rightarrow \left(\sqrt{5}x+1\right)\left(2x-\sqrt{5}\right)=0\\ \Rightarrow \sqrt{5}x+1=0or2x-\sqrt{5}=10\\ \Rightarrow x=\frac{-1}{\sqrt{5}}orx=\frac{\sqrt{5}}{2}\end{array}$