4x^2-16x=48

or, 4(x^2) - 16x - 48 = 0

or, 4[(x^2) - 4x - 12] = 0

or, [(x^2) - 4x - 12] = 0

or, [(x^2) - 6x + 2x - 12] = 0

or, [x(x-6) + 2(x-6)] = 0

or, [(x+2)*(x-6)] = 0

Thus, either, (x+2) = 0 , which gives x = -2 ...........(1)

or, (x-6) = 0 ; which gives x = 6..............(2)

### 4x^2-16x=48, solve for x

Step 1: turn it into a quadratic equation

4x^2-16x-48 = 0

Step 2: factor the polynomial by factoring out the 4s

**4**(x^2-4x-12) = 0

Now factor the trinomial that you have left

4(x+2)(x-6) = 0

If you know how to solve for a solution set, you should be able to work the rest of the problem from there.

You should get a solution set of

(-2,6)