Solve this showing your work: 6x² + 42x = 0?

6x² + 42x = 0
6x(x+7)=0
divide both sides by 6x...and you get:
x+7=0
subtract both sides by -7
and you get:
x=-7

6x (x + 7) = 0
x = 0 , x = - 7

dont forget... you can also divide both sides by (x+7)
so the answer can be zero also

Therefore
x= -7 or 0

First response is correct, but incomplete. Since x = 0 is also a solution, you have that as well as x = -7.

6x^2 + 42x = 0
(6x^2 + 42x)/6 = 0/6
x^2 + 7x = 0
x(x + 7) = 0

x = 0

x + 7 = 0
x = -7

∴ x = -7 , 0

The other answer is slightly incorrect in dividing by 6x, since that assumes that x is not 0. Since we do not know this, we should avoid that. Instead try this:

Divide both sides of the equation by 6 so that it becomes

x^2 + 7x = 0

Now, use the quadratic formula to ensure you find both roots. The quadratic formula is given here: http://www.purplemath.com/modules/quadform.htm

Applying the quadratic formula here, notice that a = 1, b = 7, and c = 0. Plugging these values into the quadratic formula, you get

x = (-7 +/- sqrt(7^2 - 4*1*0))/2

where +/- means "plus or minus".

Simplifying, you get

x = (-7 +/- 7)/2 = both (-7 + 7)/2 and (-7 - 7)/2

This means that x = 0 or x = -7. You can plug both of these back into the original formula to check that they are roots. So the answer x = -7 misses out on the fact that x = 0 is a root, and therefore you cannot divide out by 6x and recover this root successfully.