How do you solve #6x^2 + 2x = 104#?

2 Answers
Apr 1, 2018

Answer:

#x=-4, -13/3#

Explanation:

Subtract #104# from both sides to get
#6x^2+2x-104=104-104# => #6x^2+2x-104=0#.

Apply the quadratic formula.

#a=6#
#b=2#
#c=-104#

#x=(-2±√2^2-4*6*(-104))/(2*6)#

Simplify to get
#x=(-2±√2^2+2496)/12#

Simplify even more to get #x=(-2±√2500)/12#
#x=(-2±50)/12#

If #x=(-2+50)/12# then #x=-4#.
If #x=(-2-50)/12# then #x=-13/3#

Yay!

Apr 1, 2018

Answer:

4 and #- 13/3#

Explanation:

Use the new Transforming Method (Socratic, Google Search).
#6x^2 + 2x - 104 = 0#
#y = 3x^2 + x - 52 = 0#
Transformed equation:
#y' = x^2 + x - 156 = 0#
Proceeding. Find the 2 real roots of y', then, divide them by a = 3.
Find 2 real roots, that have opposite signs (ac < 0), knowing their sum (-b = -1) and their product (ac = - 156). They are 12 and - 13.
Back to y, the 2 real roots of y are:
#x1 = 12/a = 12/3 = 4#, and #x2 = - 13/a = - 13/3#