How do you solve #3x ^ { 2} + 172= 42x#?

1 Answer
Sep 25, 2017

See a solution process below:

Explanation:

First, subtract #color(red)(42x)# from each side of the equation to put the equation in standard form while keeping the equation balanced:

#3x^2 - color(red)(42x) + 172 = 42x - color(red)(42x)#

#3x^2 - 42x + 172 = 0#

Now, we can use the quadratic equation to solve this problem:

The quadratic formula states:

For #color(red)(a)x^2 + color(blue)(b)x + color(green)(c) = 0#, the values of #x# which are the solutions to the equation are given by:

#x = (-color(blue)(b) +- sqrt(color(blue)(b)^2 - (4color(red)(a)color(green)(c))))/(2 * color(red)(a))#

Substituting:

#color(red)(3)# for #color(red)(a)#

#color(blue)(-42)# for #color(blue)(b)#

#color(green)(172)# for #color(green)(c)# gives:

#x = (-color(blue)((-42)) +- sqrt(color(blue)((-42))^2 - (4 * color(red)(3) * color(green)(172))))/(2 * color(red)(3))#

#x = (42 +- sqrt(1764 - 2064))/6#

#x = (42 +- sqrt(-300))/6#

#x = (42 +- sqrt(100 * -3))/6#

#x = (42 +- sqrt(100)sqrt(-3))/6#

#x = (42 - 10sqrt(-3))/6# and #x = (42 + 10sqrt(-3))/6#

#x = 42/6 - (10sqrt(-3))/6# and #x = 42/6 + (10sqrt(-3))/6#

#x = 7 - (5sqrt(-3))/3# and #x = 7 + (5sqrt(-3))/3#