How do you solve #(x-4)(x+3) = 6#?

1 Answer
Feb 5, 2016

Answer:

#x= (1+-sqrt(73))/2#

Explanation:

How to solve #(x-4)(x+3) = 6#

Step 1: Multiple the expand and combine like terms for the left side of the equation

#(x-4)(x+3) = 6#

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

#(x^2-x+-2) = 6#

Step 2-Set the equation equal to zero by subtracting both side of the equation by 6

#x^2 -x-12= 6#

#-6 " " " -6#
#=====#
#color(red)(x^2 -x -18= 0#

Step 3 Solve the equation using quadratic formula, since we can't factor it

Recall: Quadratic formula: #x= (-b+- sqrt((b)^2-4ac))/(2a)#

#color(red)(x^2 -x -18= 0#

#a = 1, b = -1, c= -18#

Substitute into the quadratic formula, we get
#x= (-(-1)+- sqrt((-1)^2-4(1)(-18)))/(2(1)#

#x= (1+-sqrt(1+72))/(2)#

#x= (1+-sqrt(73))/2#