How do you solve #(x-4)(x+3) = 6#?
1 Answer
Feb 5, 2016
Explanation:
How to solve
Step 1: Multiple the expand and combine like terms for the left side of the equation
#(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:
#color(red)(x^2 -x -18= 0#
Substitute into the quadratic formula, we get
#x= (1+-sqrt(1+72))/(2)#
#x= (1+-sqrt(73))/2#