How do you find the zeros, real and imaginary, of #y= x^2+12x+6 # using the quadratic formula?

1 Answer
Aug 12, 2017

Answer:

See a solution process below:

Explanation:

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)(1)# for #color(red)(a)#

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

#color(green)(6)# for #color(green)(c)# gives:

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

#x = (-color(blue)(12) +- sqrt(144 - 24))/2#

#x = (-color(blue)(12) +- sqrt(120))/2#

#x = (-color(blue)(12) +- sqrt(4 * 30))/2#

#x = (-color(blue)(12) - sqrt(4 * 30))/2# and #x = (-color(blue)(12) + sqrt(4 * 30))/2#

#x = (-color(blue)(12) - sqrt(4)sqrt(30))/2# and #x = (-color(blue)(12) + sqrt(4)sqrt(30))/2#

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

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

#x = -6 - sqrt(30)# and #x = -6 + sqrt(30)#