(x-1)(x+3)=4x I cant remember where to apply the 4x

BEST ANSWER

multiply out the brackets
x^2+3x-x-3
x^2+2x-3

then take 4 to the other side so its

x^2+2x-3-4x

x^2-2x-3=0

(equate to 0 )

(x-3)(x+1)=0
x=3 or x=-1

The Captain and Watz are correct.

I have noticed that neither bothered to check their result.

I check my results all of the time. It allows me to be certain that my answers are correct. It does take some extra time, but the price is worth the peace of mind provided.

I've been doing this for over 40 years. I rarely make an error, but by checking I catch almost all of them. No one is perfect.

ProfRay

(x-1)(x+3)=4x
x^2+2x-3=4x
x^2+2x-4x-3=0
x^2-2x-3=0
(x+1)(x-3)=0
x+1=0 x-3=0
x=-1 x=3

x=-1, 3 answer//

(x - 1) * (x + 3) = 4x
x^2 - x + 3x - 3 = 4x
x^2 + 2x - 3 = 4x
x^2 - 2x - 3 = 0
(x - 3) * (x + 1) = 0
x = -1 , 3

Multiply out the left side. Then subtract 4x from both sides.
That will give you a quadratic equation, with right side 0.
Solve it.

x^2+2x-3=4x

x^2-2x-3=0
(x-3)(x+1)=0
x=3 x=-1

expand the left side, then subtract 4x from both sides