# How do you solve #2x^2=x+28 #?

##### 3 Answers

Move all values and variables to the left and either factor, solve with the quadratic equation, or solve by completing the square; you could also leave the expression alone and solve by graphing.

#### Explanation:

Next, try to factor this expression.

Where

The factors of -28 are:

Of these factors, only the last pair will allow you to make

Check your work by plugging these values back into the original expression.

#### Explanation:

You could solve this quadratic equation by completing the square.

To do that, start by getting your quadratic to the general form

#color(blue)(x^2 + b/ax = -c/a)#

by adding

Now, you need to add a term to both sides of the equation so that the left side of the equation can be written as the **square of a binomial**.

The coefficient of the **square** the result to get

Add

The left side of the equation can now be rewritten as

You now have

Take the square root of both sides of the equation

The two solutions to the equation will thus be

Solve

#### Explanation:

I use the new Transforming Method.

Transformed equation

Factor pairs of (-56) --> (-4, 14)(-7, 8). This sum is 1 = -b. Two real roots of (2) are: -7 and 8.

The 2 real roots of equation (1) are: