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

Use the improved quadratic formula (Google, Yahoo Search)
`D = d^2 = b^2 - 4ac = 4 + 20 = 24` --> `d = +- 2sqrt6`
There are 2 real roots:
`x = - 2/2 +- (2sqrt6)/2 = -1 +- sqrt6`

`color(blue)(x^2+2x-5=0`

This is a Quadratic equation (in form `ax^2+bx+c=0`)

Use Quadratic equation

`color(brown)(x=(-b+-sqrt(b^2-4ac))/(2a)`

Where

`color(red)(a=1,b=2,c=-5`

`rarrx=(-2+-sqrt(2^2-4(1)(-5)))/(2(1))`

`rarrx=(-2+-sqrt(4-4(-5)))/(2)`

`rarrx=(-2+-sqrt(4-(-20)))/(2)`

`rarrx=(-2+-sqrt(4-(-20)))/(2)`

`rarrx=(-2+-sqrt(4+20))/(2)`

`rarrx=(-2+-sqrt(24))/(2)`

`rarrx=(-2+-sqrt(4*6))/(2)`

`rarrx=(-2+-2sqrt(6))/(2)`

`rarrx=(-cancel2+-cancel2sqrt(6))/(cancel2)`

`color(green)(rArrx=-1+-sqrt6`

Another approach would be to complete the square. It is another form of the standard `x=(-b+-sqrt(b^2-4ac))/(2a)` .

`color(brown)("It has to be another form otherwise you would not be")``color(brown)("able to solve for x when y=0.")`

Given:`" "x^2+2x-5=0`

Note that standard form of `" "y=ax^2+bx+c` becomes

`y=a(x+b/(2a))^2+c+k" "` where `k` is a corrective constant

`=>y= (x+1)^2-5+k`

The error comes from `(b/2)^2 -> (+2/2)^2`

So `(+2/2)^2+k=0" "=> k=-1` giving

`y=(x+1)^2-6`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`y+6=(x+1)^2`

Taking the square root of both sides

`sqrt(y+6)=x+1`

But for this question `y=0` giving

`+-sqrt(6)-1=x`

`color(blue)(=> x~~ 1.449" or "x~~-3.449" to 3 decimal places")`