How do you solve using the completing the square method `x^2 + 3x - 10 = 0`?

Given:`" "x^2+3x-10=0`.......................(1)

`color(blue)("Determine "y_("intercept"))`

Read directly from equation (1)

`color(blue)(y_("intercept")=-10)`

`color(brown)("'~~~~~~~~~~~~~ Tip! ~~~~~~~~~~~~~~~~~~~~")`
`color(green)("As the equation is already in the form ")`

`color(green)(y=a(x^2+b/a x) + c)`

`color(green)("In this case "a=1)`

`color(green)(x_("vertex")=(-1/2)xxb/a = -3/2)`

`color(brown)("'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~")`

`color(blue)("Step 1")`
Write equation (1) as `(x^2+3x)-10=0`

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Step 2")`

Add the adjustment constant `k`
`(x^2+3x)-10+k=0`

'~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Step 3")`
Move the power from `x^2` to outside the bracket
`(x+3x)^2-10+k=0`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Step 3")`

Remove the `x" from "3x`
`(x+3)^2-10+k=0`

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Step 4")`

Multiply the 3 inside the bracket by `1/2`
`(x+3/2)^2-10+k=0`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Step 4")`
If we multiply out the bracket we end up with an additional term to those in the original equation. That term is `(3/2)^2` derived from `(?+3/2)(?+3/2)=(3/2)^2`. This term must be removed which is achieved by making `k=-(3/2)^2`

So now we have

`color(brown)((x+3/2)^2-10+k=0)color(green)(->(x+3/2)^2-10-(3/2)^2=0)`

`color(blue)("Completing the square "->)color(magenta)( (x+3/2)^2-49/4=0)`......................(2)
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`" "color(blue)("Vertex" -> (x,y)->(-3/2,-49/4))`

From equation (2)

`(x+3/2)^2=49/4`

Square rooting both sides

`x+3/2= +-sqrt(49)/sqrt(4)`

`color(blue)(x_("intercepts")=-3/2+-7/2 =`-5 or +2)#