Cubic equation and its roots

$Cubic\quad Equation\\ f(x)=ax^{ 3 }+bx^{ 2 }+cx+d\\ setting\quad f\left( x \right) =0\\ ax^{ 3 }+bx^{ 2 }+cx+d=0\\ a\neq 0\\ Graph\quad of\quad Cubic\quad equation\quad Is\quad shown\quad above\\ Roots\quad of\quad cubic\quad equation\\ the\quad nature\quad of\quad roots\\ \Delta =18abcd-4b^{ 3 }d+b^{ 2 }c^{ 2 }-4ac^{ 3 }-27a^{ 2 }d^{ 2 }.\, \\ IfΔ>0,then\quad the\quad equation\quad has\quad three\quad distinct\quad real\quad roots.\\ IfΔ=0,then\quad the\quad equation\quad has\quad a\quad multiple\quad root\quad and\quad all\quad its\quad roots\quad are\quad real.\\ IfΔ<0,then\quad the\quad equation\quad has\quad one\quad real\quad root\quad and\quad two\quad non\quad real\quad complex\quad conjugate\quad roots.\\ General\quad formula\quad for\quad roots\\ x_{ k }=-\frac { 1 }{ 3a } \left( b+u_{ k }C+\frac { \Delta _{ 0 } }{ u_{ k }C } \right) ,\\ \qquad k\in \{ 1,2,3\} \\ whereu_{ 1 }=1,\qquad u_{ 2 }={ \frac { -1+i\sqrt { 3 } }{ 2 } },\qquad u_{ 3 }={ \frac { -1-i\sqrt { 3 } }{ 2 } }\quad \\ u_{ 1 },\quad u_{ 2,\quad }u_{ 3 }\quad are\quad the\quad three\quad cube\quad roots\quad of\quad unity,\\ and\quad whereC=\sqrt [ 3 ]{ \frac { \Delta _{ 1 }+\sqrt { \Delta _{ 1 }^{ 2 }-4\Delta _{ 0 }^{ 3 } } }{ 2 } } \qquad \\ with\Delta _{ 0 }=b^{ 2 }-3ac\\ \Delta _{ 1 }=2b^{ 3 }-9abc+27a^{ 2 }d\\ and\quad \Delta _{ 1 }^{ 2 }-4\Delta _{ 0 }^{ 3 }=-27\, a^{ 2 }\, \Delta ,\quad where\Delta \quad is\quad the\quad discriminant\quad discussed\quad above.\\ Special\quad cases\\ If\quad \Delta \neq 0\quad and\quad \Delta _{ 0 }=0,\\ the\quad sign\quad of\quad \sqrt { \Delta _{ 1 }^{ 2 }-4\Delta _{ 0 }^{ 3 } } =\sqrt { \Delta _{ 1 }^{ 2 } } \quad has\quad to\quad be\quad chosento\quad have\quad C\neq 0,\quad that\quad is\quad one\quad should\quad define\quad \sqrt { \Delta _{ 1 }^{ 2 } } =\Delta _{ 1 },\quad which\quad ever\quad is\quad the\quad sign\quad of\quad \Delta _{ 1 }.\quad \\ If\quad \Delta =0\quad and\quad \Delta _{ 0 }=0,\quad the\quad three\quad roots\quad are\quad equal:x_{ 1 }=x_{ 2 }=x_{ 3 }=-\frac { b }{ 3a } .\quad \\ If\Delta =0and\Delta _{ 0 }\neq 0,the\quad above\quad expression\quad for\quad the\quad roots\quad is\quad correct\quad but\quad misleading,\quad hiding\quad the\quad fact\quad that\quad no\quad radical\quad is\quad needed\quad to\quad represent\quad the\quad roots.\\ In\quad fact,\quad in\quad this\quad case,\quad there\quad is\quad a\quad double\quad root\quad ,x_{ 1 }=x_{ 2 }=\frac { 9ad-bc }{ 2\Delta _{ 0 } } ,and\quad a\quad simple\quad root\quad x_{ 3 }=\frac { 4abc-9a^{ 2 }d-b^{ 3 } }{ a\Delta _{ 0 } } .$