Đáp án: $ x=\dfrac{-7\pm\sqrt{29}}{2}$
Giải thích các bước giải:
Ta có:
$(x+1)(2-x)-(5x+5)(x+2)=-4x^2+2$
$\to (2x-x^2+2-x)-(5x^2+10x+5x+10)=-4x^2+2$
$\to 2x-x^2+2-x-5x^2-10x-5x-10=-4x^2+2$
$\to -6x^2-14x-8=-4x^2+2$
$\to 2x^2+14x+10=0$
$\to x^2+7x+5=0$
$\to x^2+2x\cdot\dfrac72+(\dfrac72)^2+5=(\dfrac72)^2$
$\to (x+\dfrac72)^2+5=(\dfrac72)^2$
$\to (x+\dfrac72)^2=\dfrac{29}{4}$
$\to x+\dfrac72=\pm\dfrac{\sqrt{29}}{2}$
$\to x=\dfrac{-7\pm\sqrt{29}}{2}$
