0\\\\x_1=12\; ,\; x_2=4\; \; (teorema\; Vieta)\\\\(x-4)(x-12)>0\; \; \; \; +++(4)---(12)+++\\\\x\in (-\infty ,4)\cup(12,+\infty )\\\\\\2)\; \; 2x-4x^2\leq 0\\\\-2x(2x-1)\leq 0\; \; \to \; \; \; 2x(2x-1)\geq 0\; \; ,\; \; x_1=0\; ,\; x_2=\frac{1}{2}\\\\+++[\, 0\, ]---[\, \frac{1}{2}\, ]+++\qquad x\in (-\infty ,0\, ]\cup [\, \frac{1}{2}\, ,+\infty )\\\\\\3)\; \; x-0,5x^2-10<0\, |\cdot (-2)\\\\x^2-2x+20>0\; \; ,\; \; D/4=1-20=-19<0\; \; \to \; \; x\in (-\infty ,+\infty )" alt="1)\; \; x^2-16x+48>0\\\\x_1=12\; ,\; x_2=4\; \; (teorema\; Vieta)\\\\(x-4)(x-12)>0\; \; \; \; +++(4)---(12)+++\\\\x\in (-\infty ,4)\cup(12,+\infty )\\\\\\2)\; \; 2x-4x^2\leq 0\\\\-2x(2x-1)\leq 0\; \; \to \; \; \; 2x(2x-1)\geq 0\; \; ,\; \; x_1=0\; ,\; x_2=\frac{1}{2}\\\\+++[\, 0\, ]---[\, \frac{1}{2}\, ]+++\qquad x\in (-\infty ,0\, ]\cup [\, \frac{1}{2}\, ,+\infty )\\\\\\3)\; \; x-0,5x^2-10<0\, |\cdot (-2)\\\\x^2-2x+20>0\; \; ,\; \; D/4=1-20=-19<0\; \; \to \; \; x\in (-\infty ,+\infty )" align="absmiddle" class="latex-formula">