Đáp án: $x=\dfrac{-21\pm\sqrt{537}}4$

Giải thích các bước giải:

$x\cdot 2x-2+2x\cdot 11-x=10$

$\to 2x^2-2+22x-x=10$

$\to 2x^2+21x-2=10$

$\to 2x^2+21x-12=0$

$\to 2(x+\dfrac{21}4)^2-\dfrac{537}{8}=0$

$\to 2(x+\dfrac{21}4)^2=\dfrac{537}{8}$

$\to (x+\dfrac{21}4)^2=\dfrac{537}{16}$

$\to x+\dfrac{21}4=\dfrac{\pm\sqrt{537}}4$

$\to x=\dfrac{-21\pm\sqrt{537}}4$