0\\13x^2+12x=0\\x=0\ x=-\frac{12}{13}\\\\////+///(-\frac{12}{13})\_\_-\_\_(0)////+///>x\\x\in(-\infty;-\frac{12}{13})\cup(0;+\infty)" alt="3x^2+4x<16x(x+1)\\3x^2+4x<16x^2+16x\\13x^2+12x>0\\13x^2+12x=0\\x=0\ x=-\frac{12}{13}\\\\////+///(-\frac{12}{13})\_\_-\_\_(0)////+///>x\\x\in(-\infty;-\frac{12}{13})\cup(0;+\infty)" align="absmiddle" class="latex-formula">