How do you solve #6x^2-12x+1=0# using the quadratic formula?

Question

7155 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-08-24T19:23:57

See a solution process below:

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)(6)# for #color(red)(a)#

#color(blue)(-12)# for #color(blue)(b)#

#color(green)(1)# for #color(green)(c)# gives:

#x = (-color(blue)((-12)) +- sqrt(color(blue)((-12))^2 - (4 * color(red)(6) * color(green)(1))))/(2 * color(red)(6))#

#x = (color(blue)(12) +- sqrt(144 - 24))/12#

#x = (color(blue)(12) +- sqrt(120))/12#

#x = (color(blue)(12) +- sqrt(4 * 30))/12#

#x = (color(blue)(12) +- 2sqrt(30))/12#

Or

#x = 1 +- sqrt(30)/6#