0\\\\D/4=12^2-16\cdot 9=0\\\\(4x-3)^2>0\\\\(4x-3)^2=0\; \; pri \; \; x=\frac{3}{4}\\\\+ + + + + + \; (\frac{3}{4})\; + + + + + +\\\\x\in (-\infty,\frac{3}{4})U(\frac{3}{4},+\infty)" alt="16x^2-24x+9>0\\\\D/4=12^2-16\cdot 9=0\\\\(4x-3)^2>0\\\\(4x-3)^2=0\; \; pri \; \; x=\frac{3}{4}\\\\+ + + + + + \; (\frac{3}{4})\; + + + + + +\\\\x\in (-\infty,\frac{3}{4})U(\frac{3}{4},+\infty)" align="absmiddle" class="latex-formula">