[tex]14x^2 +11x = 15\ \ |\ subtract\ 15 \ to\ both\ sides \\\\ 14x^2 +11x-15 =0\\\\a=14, \ \ \ b=11, \ \ \ c=-15 \\\\ x_{1}=\frac{-b-\sqrt{b^2-4ac}}{2a}=\frac{-11-\sqrt{11^2-4 \cdot 14 \cdot (-15) }}{2\cdot 14}=\frac{-11-\sqrt{121+840 }}{28}=\\\\=\frac{-11-\sqrt{961 }}{28}=\frac{-11-31}{28}=\frac{-42}{28}=-\frac{3}{2}[/tex]
[tex]x_{2}=\frac{-b+\sqrt{b^2-4ac}}{2a}=\frac{-11+\sqrt{11^2-4 \cdot 14 \cdot (-15) }}{2\cdot 14}=\\\\=\frac{-11+31}{28}=\frac{20}{28}=\frac{5}{7}\\ \\Answer: \ x=-\frac{3}{2}\ \ \ or \ \ \ x=\frac{5}{7}[/tex]
