`a=2;c=\sqrt{5}-2`
Ta có:
`\qquad 2x^2+x(2c-\sqrt{a})-c\sqrt{2}=0`
`<=>2x^2+x(2c-\sqrt{2})-c\sqrt{2}=0`
`<=>2x^2+2cx-x\sqrt{2}-c\sqrt{2}=0`
`<=>2x(x+c)-\sqrt{2}(x+c)=0`
`<=>(x+c)(2x-\sqrt{2})=0`
$⇔\left[\begin{array}{l}x+c=0\\2x-\sqrt{2}=0\end{array}\right.$
$⇔\left[\begin{array}{l}x=-c\\2x=\sqrt{2}\end{array}\right.$
$⇔\left[\begin{array}{l}x=2-\sqrt{5}\\x=\dfrac{\sqrt{2}}{2}\end{array}\right.$
Vậy `x=2-\sqrt{5}` hoặc `x={\sqrt{2}}/2`
