Answer:
[tex]\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}[/tex]
[tex]x=\frac{3+\sqrt{29}}{2},\:x=\frac{3-\sqrt{29}}{2}[/tex]
Step-by-step explanation:
Given the equation
[tex]x^2\:-3x\:-\:5\:=\:0[/tex]
solving with the quadratic formula
[tex]\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}[/tex]
[tex]x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
[tex]\mathrm{For\:}\quad a=1,\:b=-3,\:c=-5[/tex]
[tex]x_{1,\:2}=\frac{-\left(-3\right)\pm \sqrt{\left(-3\right)^2-4\cdot \:1\cdot \left(-5\right)}}{2\cdot \:1}[/tex]
[tex]x_{1,\:2}=\frac{-\left(-3\right)\pm \sqrt{29}}{2\cdot \:1}[/tex]
separating the solutions
[tex]x_1=\frac{-\left(-3\right)+\sqrt{29}}{2\cdot \:1},\:x_2=\frac{-\left(-3\right)-\sqrt{29}}{2\cdot \:1}[/tex]
solving
[tex]x=\frac{-\left(-3\right)+\sqrt{29}}{2\cdot \:\:1}[/tex]
[tex]=\frac{3+\sqrt{29}}{2\cdot \:1}[/tex]
[tex]=\frac{3+\sqrt{29}}{2}[/tex]
also solving
[tex]\:x=\frac{-\left(-3\right)-\sqrt{29}}{2\cdot \:\:1}[/tex]
[tex]=\frac{3-\sqrt{29}}{2\cdot \:1}[/tex]
[tex]=\frac{3-\sqrt{29}}{2}[/tex]
[tex]\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}[/tex]
[tex]x=\frac{3+\sqrt{29}}{2},\:x=\frac{3-\sqrt{29}}{2}[/tex]
