How do you solve #92- 42x = 2x^2#?

1 Answer
May 2, 2018

Answer:

#x = 2 or x = -23#

Explanation:

Notice that it is a quadratic equation because there is an #x^2# term. The first step is to move all the terms to one side to make the equation equal to #0#

#2x^2 +42x-92=0" "larr div 2#

#x^2 +21x-46=0" "larr# find factors if possible

#(x+23)(x-2)=0#

Either factor could be equal to #0#

#x+23 = 0 " "rarr x =-23#

#x-2=0" "rarr x = 2#