Odpowiedź :
c)
[tex]12-5x<2x^2\\2x^2+5x-12>0\\\Delta=b^2-4ac=5^2-4*2*\left(-12\right)=25+96=121\\x_1=\cfrac{-b-\sqrt\Delta}{2a}=\cfrac{-5-\sqrt{121}}{2*2}=\cfrac{-5-11}{4}=-4\\x_2=\cfrac{-b+\sqrt\Delta}{2a}=\cfrac{-5+\sqrt{121}}{2*2}=\cfrac{-5+11}{4}=1.5\\\\x\in\mathbb{R}\setminus\langle-4;1.5\rangle\text{ lub }x\in(-\infty;-4)\cup(1.5;\infty)[/tex]
d)
[tex](3x-1)^2-16\geqslant(2x-4)(2x+4)\\9x^2-6x+1-16\geqslant 4x^2-16\\5x^2-6x+1\geqslant 0\\\Delta=b^2-4ac=\left(-6\right)^2-4*5*1=36-20=16\\x_1=\cfrac{-b-\sqrt\Delta}{2a}=\cfrac{6-\sqrt{16}}{2*5}=\cfrac{6-4}{10}=0.2\\x_2=\cfrac{-b+\sqrt\Delta}{2a}=\cfrac{6+\sqrt{16}}{2*5}=\cfrac{6+4}{10}=1\\\\x\in\mathbb{R}\setminus(0.2;1)\text{ lub }x\in(-\infty;0.2\rangle\cup\langle1;\infty)[/tex]