How do you find the zeros, real and imaginary, of #y=-x^2-12x+11# using the quadratic formula?

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Answer 1 · 2018-01-08T12:58:32

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

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

#color(green)(11)# for #color(green)(c)# gives:

#x = (-color(blue)(-12) +- sqrt(color(blue)(-12)^2 - (4 * color(red)(-1) * color(green)(11))))/(2 * color(red)(-1))#

#x = (12 +- sqrt(144 - (-44)))/(-2)#

#x = (12 +- sqrt(144 + 44))/(-2)#

#x = (12 +- sqrt(188))/(-2)#

#x = (12 +- sqrt(4 xx 47))/(-2)#

#x = (12 +- sqrt(4)sqrt(47))/(-2)#

#x = (12 +- 2sqrt(47))/(-2)#

#x = -6 +- sqrt(47)#