Question

Use the quadratic formula to solve x^2 -3x - 5 = 0. Round to two decimal places.

Answer 1

`\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}`

`x=(3+√(29))/(2),\:x=(3-√(29))/(2)`

Explanation:

Given the equation

`x^2\:-3x\:-\:5\:=\:0`

solving with the quadratic formula

`\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}`

`x_(1,\:2)=(-b\pm √(b^2-4ac))/(2a)`

`\mathrm{For\:}\quad a=1,\:b=-3,\:c=-5`

`x_(1,\:2)=(-\left(-3\right)\pm √(\left(-3\right)^2-4\cdot \:1\cdot \left(-5\right)))/(2\cdot \:1)`

`x_(1,\:2)=(-\left(-3\right)\pm √(29))/(2\cdot \:1)`

separating the solutions

`x_1=(-\left(-3\right)+√(29))/(2\cdot \:1),\:x_2=(-\left(-3\right)-√(29))/(2\cdot \:1)`

solving

`x=(-\left(-3\right)+√(29))/(2\cdot \:\:1)`

`=(3+√(29))/(2\cdot \:1)`

`=(3+√(29))/(2)`

also solving

`\:x=(-\left(-3\right)-√(29))/(2\cdot \:\:1)`

`=(3-√(29))/(2\cdot \:1)`

`=(3-√(29))/(2)`

`\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}`

`x=(3+√(29))/(2),\:x=(3-√(29))/(2)`