Question

How do you solve `0 = x^2 + 10x + 5`?

Answer 1

`x = -5+-2sqrt(5)`

Explanation:

Complete the square and use the difference of squares identity:

`a^2-b^2=(a-b)(a+b)`

with `a=(x+5)` and `b=2sqrt(5)` as follows:

`0 = x^2+10x+5`

`=(x+5)^2-25+5`

`=(x+5)^2-20`

`=(x+5)^2-(2sqrt(5))^2`

`=((x+5)-2sqrt(5))((x+5)+2sqrt(5))`

`=(x+5-2sqrt(5))(x+5+2sqrt(5))`

Hence:

`x = -5+-2sqrt(5)`

Answer 2

`x=-5+2sqrt(5)color(white)("XXX")orcolor(white)("XXX")x=-5-2sqrt(5)`

Explanation:

The easiest way for this particular example is to use the quadratic formula:

For an equation of the form `color(red)(a)x^2+color(blue)(b)x+color(green)(c)=0`
the solutions are given by `x=(-color(blue)(b)+-sqrt(color(blue)(b)^2-4color(red)(a)color(green)(c)))/(2color(red)(a))`

For the given example

`color(red)(a)=1`
`color(blue)b=10` and
`color(green)c=5`

So
`color(white)("XXX")x=(-10+-sqrt(10^2-4(1)(5)))/(2(1))`

`color(white)("XXXX")=(-10+-4sqrt(5))/2`

`color(white)("XXXX")=-5+-2sqrt(5)`