Question

Write the quadratic equation in general quadratic form below.
3x^2 + 6x = 12

Answer 1

This is similar to the one that I showed you before, so what would you do here to start?

Answer 2

3x^2 + 6x - 12 = 0
\[x = (-b \pm \sqrt{b ^{2}-4ac})\div 2a\]

Answer 3

the thing is I only have certain options so not all answers work even when I work the problem out..

Answer 4

hmm not sure what to make of the pic, but all you're doing is subtracting 12 from both sides to go from

3x^2 + 6x = 12

to

3x^2 + 6x - 12 = 0

and that's it (because it's now in standard form)


Unless you wanted to solve for x?

Answer 5

yes, I have to solve for x

Answer 6

oh ok, thanks for clarifying

Answer 7

3x^2 + 6x = 12
3x^2 + 6x -12

Answer 8

To solve for x, let's use the quadratic formula


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


x = (-(6)+-sqrt((6)^2-4(3)(-12)))/(2(3))


x = (-6+-sqrt(36-(-144)))/(6)


x = (-6+-sqrt(180))/6


x = (-6+sqrt(180))/6 or x = (-6-sqrt(180))/6


x = (-6+6*sqrt(5))/6 or x = (-6-6*sqrt(5))/6


x = -1+sqrt(5) or x = -1-sqrt(5)


So the solutions are x = -1+sqrt(5) or x = -1-sqrt(5)