(3х-2)(х +3)≥2х(2 степінь)+ 12

[latex](3x-2)( x^{2} x+3) \geq 2 x^{2} +12 \\ 3 x^{2} +9x-2x-6 \geq 2 x^{2} +12\\ 3 x^{2} +7x-6-2 x^{2} -12 \geq 0 \\ x^{2} +7x-18 \geq 0 \\ x^{2} +7x-18=0\\ D=49+72=121 \\ \sqrt{D} =11 \\ x_{1} = \frac{-7+11}{2} =2 \\ x_{2} = \frac{-7-11}{2}=-9\\ \\ x\in (-\infty ;-9]\cup[2;+ \infty) [/latex]