$(2x+3)^2 -(5x-4)=(x+5)^2 -(3x-1)(7x+2)-(x^2 -1+1)$
$\Leftrightarrow (4x^2 + 12x + 9) - 5x + 4 = (x^2 + 10x + 25) - (21x^2 + 6x - 7x - 2) - x^2$
$\Leftrightarrow 4x^2 + 7x + 13 = x^2 + 10x + 25 - (21x^2 - x - 2) - x^2$
$\Leftrightarrow 4x^2 + 7x + 13 = x^2 + 10x + 25 - 21x^2 + x + 2 - x^2$
$\Leftrightarrow 4x^2 + 7x + 13 = (1 - 21 - 1)x^2 + (10 + 1)x + (25 + 2)$
$\Leftrightarrow 4x^2 + 7x + 13 = -21x^2 + 11x + 27$
$\Leftrightarrow 4x^2 + 21x^2 + 7x - 11x + 13 - 27 = 0$
$\Leftrightarrow 25x^2 - 4x - 14 = 0$
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$\Delta = (-4)^2 - 4 \cdot 25 \cdot (-14)$
$\Delta = 16 + 1400$
$\Delta = 1416$
$\sqrt{\Delta} = \sqrt{1416} = \sqrt{4 \cdot 354} = 2\sqrt{354}$
$x = \dfrac{-(-4) \pm 2\sqrt{354}}{2 \cdot 25}$
$x = \dfrac{4 \pm 2\sqrt{354}}{50}$
$x = \dfrac{2(2 \pm \sqrt{354})}{50}$
$x = \dfrac{2 \pm \sqrt{354}}{25}$
Vậy pt có hai nghiệm là $x_1 = \dfrac{2 + \sqrt{354}}{25}$ và $x_2 = \dfrac{2 - \sqrt{354}}{25}$