How do you find the zeros, real and imaginary, of #y= -3x^2-2x-18 #using the quadratic formula?

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Answer 1 · 2017-08-21T16:27:21

See a solution process below:

To find the zeroes we set #y# to #0# giving:

#0 = -3x^2 - 2x - 18#

We can now use the quadratic equation to solve this problem:

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

#color(green)(-18)# for #color(green)(c)# gives:

#x = (-color(blue)((-2)) +- sqrt(color(blue)((-2))^2 - (4 * color(red)(-3) * color(green)(-18))))/(2 * color(red)(-3))#

#x = (2 +- sqrt(4 - (216)))/(-6)#

#x = (2 +- sqrt(-212))/(-6)#

#x = (2 +- sqrt(4 * -53))/(-6)#

#x = (2 +- sqrt(4)sqrt(-53))/(-6)#

#x = (2 +- 2sqrt(-53))/(-6)#

#x = (1 +- 1sqrt(-53))/(-3)#

#x = (1 +- sqrt(-53))/(-3)#