Question

How do you solve using the quadratic formula `3x^2 + 6x = 12`?

Answer 1

`-1 +- sqrt5`

Explanation:

First get the 12 to the other side.
`3x^2+6x-12=0`

Quadratic Formula is
`(-b +- sqrt(b^2-4ac))/"2a"`

where
a=3
b=6
c=-12

Now we just plug it in
`(-(6)+-sqrt(6^2-4(3)(-12)))/"2(3)"`

You can split this up like so
`-6/"2(3)" +-sqrt(6^2-4(3)(-12))/"2(3)"`

Then get this
`-6/"6" +- sqrt(36-(-144))/"6"`

Simplify
`-1 +- sqrt(180)/"6"`

Then we simplify the square-root like so,
`sqrt(36)*sqrt(5)=sqrt(180)`

Thus
`6sqrt(5)=sqrt(180)`

`-1 +-(6sqrt(5))/"6"`

Which simplifies to
`-1 +-sqrt(5)`