Odpowiedź:
[tex](x-2)^2\leq (2x+3)^2-x+7\\\\x^2-4x+4\leq 4x^2+12x+9-x+7\\\\x^2-4x+4\leq 4x^2+11x+16\\\\x^2-4x+4-4x^2-11x-16\leq 0\\\\-3x^2-15x-12\leq 0\ \ /:(-3)\\\\x^2+5x+4\geq 0\\\\a=1\ \ ,\ \ b=5\ \ ,\ \ c=4\\\\\Delta=b^2-4ac\\\\\Delta=5^2-4\cdot1\cdot4=25-16=9\\\\\sqrt{\Delta}=\sqrt{9}=3\\\\x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-5-3}{2\cdot1}=\frac{-8}{2}=-4\\\\x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-5+3}{2\cdot1}=\frac{-2}{2}-1[/tex]
