x\\x\in(-4;1)" alt="\sf x^2+3x-4<0\\y=x^2+3x-4;y=0\\x^2+3x-4=0\\D=b^2-4ac=9+4\cdot 4=25=5^2\\x_{1,2}=\frac{-b\pm \sqrt{D}}{2a}=\frac{-3\pm 5}{2}=\left |{ {{1} \atop {-4}} \right. \\+++(-4)---(1)+++>x\\x\in(-4;1)" align="absmiddle" class="latex-formula">
Ответ: 