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

Question

1777 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-08T18:50:32

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)(-12)# for #color(red)(a)#

#color(blue)(-2)# for #color(blue)(b)#

#color(green)(-12)# for #color(green)(c)# gives:

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

#x = (color(blue)(2) +- sqrt(color(blue)(4 - 576)))/-24#

#x = -(2 +- sqrt(-572))/24#

#x = -(2 +- sqrt(4 * -143))/24#

#x = -(2 +- sqrt(4)sqrt(-143))/24#

#x = -(2 +- 2sqrt(-143))/24#

#x = -(1 +- 1sqrt(-143))/12#

#x = -(1 +- sqrt(-143))/12#