Answer:
[tex]x=2+\sqrt{11},\:x=2-\sqrt{11}[/tex]
Step-by-step explanation:
[tex]x^2-4x-7=0\\\mathrm{Solve\:with\:the\:quadratic\:formula}\\Quadratic\:Equation\:Formula\\\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\\mathrm{For\:}\quad a=1,\:b=-4,\:c=-7:\quad x_{1,\:2}=\frac{-\left(-4\right)\pm \sqrt{\left(-4\right)^2-4\cdot \:1\left(-7\right)}}{2\cdot \:1}\\x=\frac{-\left(-4\right)+\sqrt{\left(-4\right)^2-4\cdot \:1\left(-7\right)}}{2\cdot \:1}:\quad 2+\sqrt{11}[/tex]
[tex]x=\frac{-\left(-4\right)-\sqrt{\left(-4\right)^2-4\cdot \:1\left(-7\right)}}{2\cdot \:1}:\quad 2-\sqrt{11}\\\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}\\x=2+\sqrt{11},\:x=2-\sqrt{11}[/tex]
