How do you solve `x^2-5x-10=0` by completing the square?

`ax^2+bx+c=0`
`x^2-5x-10=0`

`a=1`
`=>`

`(x+x_1)(x+x_2)=0`
`=>`
`(x_1+x_2)=-5=b`
`(x_1*x_2)=-10=c`

`D{-10}=(-1,10),(1,-10),(-2,5),(2,-5)`

`(-1,10):`
`-1+10=9!=-5`

`(1,-10):`
`1-10=-9!=-5`

`(-2,5):`
`-2+5=3!=-5`

`(2,-5):`
`2-5=-3!=-5`

`=>`
In that case we don't have any `ZZ` to work with, so we will move to Plan B:

`x_(1,2)={-b+-sqrt(b^2-4*a*c)}/{2*a}`
`=>`

`x_(1,2)={-(-5)+-sqrt((-5)^2+4*(1)*(-10))}/{2*(1)}=`
`={5+-sqrt(25+40)}/{2}=`
`={5+-sqrt(65)}/{2}=`
`=>`
`x_1=(5+sqrt65)/2~~6.53`
`x_2=(5-sqrt65)/2~~-1.53`

`=>`
"By the book":
`(x-(5+sqrt65)/2)(x-(5-sqrt65)/2)=0`

If we check we can easily see:

`(-(5+sqrt65)/2)*(-(5-sqrt65)/2)=(25+5sqrt65-5sqrt65-65)/4=-40/4=-10=c`

`(-(5+sqrt65)/2)+(-(5-sqrt65)/2)=(-5+sqrt65--5-sqrt65)/2=-10/2=-5=b`

`color(blue)"to complete the square"`

`• " ensure the coefficient of the "x^2" term is 1 which it is"`

`• " add "(1/2"coefficient of the x-term")^2" to both sides"`

`rArrx^2+2(-5/2)xcolor(red)(+25/4)-10=0color(red)(+25/4)`

`rArr(x-5/2)^2=25/4+10=65/4`

`color(blue)"take the square root of both sides"`

`rArr(sqrt((x-5/2)^2)=+-sqrt(65/4)larrcolor(blue)"note plus or minus"`

`rArrx-5/2=+-sqrt65/2`

`rArrx=5/2+-sqrt65/2`