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

1 Answer
Mar 14, 2018

Answer:

#-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)#