How do you solve #3x^2 + 18x - 12 = 0#?

2 Answers
Nov 26, 2015

Answer:

The solutions are
#color(blue)(x =(-9+sqrt(117))/3#

#color(blue)(x =(-9-sqrt(117))/3#

Explanation:

#3x^2+18x-12=0#

The equation is of the form #color(blue)(ax^2+bx+c=0# where:
#a=3, b=18, c=-12#

The Discriminant is given by:
#color(blue)(Delta=b^2-4*a*c#

# = (18)^2-(4*3*(-12))#

# = 324 +144=468#

The solutions are found using the formula:
#color(blue)(x=(-b+-sqrtDelta)/(2*a)#

#x = ((-18)+-sqrt(468))/(2*3) = (-18+-sqrt(468))/6#

(note:#sqrt468=sqrt(2*2*117)=2sqrt117#)

#x =(-18+-2sqrt(117))/6#

#x =(cancel2*(-9+-sqrt(117)))/cancel6#

#x =(-9+-sqrt(117))/3#

The solutions are
#color(blue)(x =(-9+sqrt(117))/3#

#color(blue)(x =(-9-sqrt(117))/3#

Nov 26, 2015

Answer:

#x=-3+sqrt13#

#x=-3-sqrt13#

Explanation:

#3x^2+18x-12=0#

Divide both sides by #3#.

#x^2+6x-4=0# is a quadratic equation, #ax^2+bx+c#, where

#a=1, b=6, c=-4#

Solve by completing the square.

Add #4# to both sides of the equation.

#x^2+6x=4#

Divide #b# by #2# and square the result. Add to both sides of the equation.

#(6/2)^2=9#

#x^2+6x+9=4+9#

#x^2+6x+9=13#

Write the trinomial as a perfect square.

#(x+3)^2=13#

Take the square root of both sides.

#x+3=+-sqrt13#

Subtract #3# from both sides.

#x=-3+-sqrt13#

Solve for #x#.

#x=-3+sqrt13#

#x=-3-sqrt13#