Решите неравенства 3х(х+2)-(4-х )(4+х) больше или равно 5(х^2+1)-4(1,5х-1)

$3x(x+2)-(4-x )(4+x) \geq 5(x^2+1)-4(1.5x-1)$
$3x^2+6x-(16-x^2) \geq 5x^2+5-6x+4$
$3x^2+6x-16+x^2 \geq 5x^2-6x+9$
$4x^2+6x-16 - 5x^2+6x-9 \geq 0$
$- x^2+12x-25 \geq 0$
$x^2-12x+25 \leq 0$
$x^2-12x+25=0$
$D=(-12)^2-4*1*25=144-100=44$
$x_1= \frac{12+2 \sqrt{11} }{2} =6+ \sqrt{11}$
$x_2= \frac{12-2 \sqrt{11} }{2} =6-\sqrt{11}$

----+----[6-√11]----- - -------[6+√11]-----+-----
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Ответ: $[6- \sqrt{11} ; 6+ \sqrt{11} ]$