How to solve completing the square? 2x^2-8x-15=0

`2x^2-8x-15=0`

Completing square method:

• Separate variable terms from constant term, rearrange the equation:

`2x^2-8x=15`

• Make sure the coefficient of `x^2` is always 1.
  Divide the equation by 2:

`x^2-4x=7.5`

• Add 4 to left, completing square.

`x^2-4x+4=11.5`

• Factor the expression on the left

`(x-2)^2=11.5`

• Take the square root

`sqrt((x-2)^2)=± sqrt(11.5)`

`x-2=± sqrt11.5`

`x=± sqrt(11.5)+2` or `x=± sqrt(23/2)+2`

`2x^2-8x-15=0`

As we are completing the square of more than one `x^2`, it is best to move the constant (15) to the other side. It's sign therefore, changes - (15 not -15).

`2x^2-8x=15`

Now we divide through by two, to obtain a single `x^2`

`x^2-4x=7.5`

To complete the square, the general steps are to take half the coefficient of x. In this case, the coefficient is 4 therefore half is two. We form brackets, leaving:

`(x-2)^2`

But, if we multiplied this out we would end up with `x^2-4x+4`
We don't want this 'extra' 4, so to complete the square, we must SUBTRACT 4, leaving;

`(x-2)^2-4=7.5`

Now we solve like a standard linear equation;
`(x-2)^2=7.5+4`
`(x-2)^2=11.5`
`x-2=+-sqrt(11.5)`
`x=2+-sqrt(11.5)`

Remember: when you move across the equals sign, you carry out the opposite operation
i.e square, square root
add, subtract
multiply, divide.

Also, when you square root a number you get both a positive AND negative number.

Hope this helps!