Odpowiedź:
[tex]4x^2-11x-15\leq 0\\\\a=4\ \ ,\ \ b=-11\ \ ,\ \ c=-15\\\\\Delta=b^2-4ac\\\\\Delta=(-11)^2-4\cdot4\cdot(-15)=121+240=361\\\\\sqrt{\Delta}=\sqrt{361}=19\\\\x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-19}{2\cdot4}=\frac{11-19}{8}=\frac{-8}{8}=-1\\\\x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+19}{2\cdot4}=\frac{11+19}{8}=\frac{30}{8}=\frac{15}{4}=3\frac{3}{4}[/tex]
