$2x^{2}+x-2=0<=>(√2x)^{2}+2.√2x.\frac{1}{√2}+\frac{1}{8}-\frac{17}{8}=0$
$<=>(√2x+\frac{1}{2√2})^{2}=√\frac{17}{8}$
\(\left[ \begin{array}{l}√2+\frac{1}{2√2}=√\frac{17}{8}\\√2+\frac{1}{2√2}=-√\frac{17}{8}\end{array} \right.\)
$<=>\left[ \begin{array}{l}\frac{-1+√17}{4}\\\frac{-1-√17}{4}\end{array} \right.$
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