`2x^2-5x = 6`

If we are going to solve for x by using completing the square method, the steps are:

> Divide both sides by 2, to make the coefficient of `x^2` equal to 1.

`(2x^2-5x) /2 = 6/2`

`x^2 - 5/2x = 3`

> Divide coefficient of `x ` by 2. Then, take the square of it.

`-5/2-:2 =-5/2 * 1/2= -5/4 ` ===> `(-5/4)^2 = 25/16`

> Add both sides of the equation by `25/16` .

`x^2 -5/2x + 25/4 = 3 +25/16`

`x^2 - 5/2x +25/4 = 48/16 + 25/16`

`x^2 - 5/2x + 25/4 = 73/16`

> Then, factor left side.

`(x - 5/4)(x-5/4) = 73/16`

`(x-5/4)^2 = 73/16`

> Take the square root of both sides.

`sqrt ((x-5/4)^2) = sqrt(73/16)`

` x-5/4 = +- sqrt73/4`

`x =5/4+-sqrt73/4`

`x_1 = 5/4+sqrt73/4 = 3.386`

`x_2 = 5/4-sqrt73/4 = -0.886`

**Hence, the values of x that satisfy the equation `x^2-5x = 6` are `x_1 = 3.386` and `x_2 = -0.886` .**

`2x^2-5x = 6`

`2x^2-5x-6 = 0`

We know the roots of `ax^2+bx+c = 0` can be written as;

x = `[-b+-sqrt(b^2-4ac)]/(2a)`

For our question;

a=2

b=-5

c=-6

x = `[-(-5)+-sqrt((-5)^2-4*2*(-6))]/(2xx2)`

So the roots are;

x1 = `[-(-5)+sqrt((-5)^2-4*2*(-6))]/(2xx2)` = 3.386

x2 = `[-(-5)-sqrt((-5)^2-4*2*(-6))]/(2xx2)` = -0.886

**So the roots of the `2x^2-5x` = 6 equation is x= 3.386 and x=-0.886.**