How do you find the zeros, real and imaginary, of #y=-2x^2-19x+42# using the quadratic formula?

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Answer 1 · 2017-10-21T12:36:55

See a solution process below:

To find the zeros of a quadratic equation you set the quadratic equal to #0#:

#0 = -2x^2 - 19x + 42#

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)(-2)# for #color(red)(a)#

#color(blue)(-19)# for #color(blue)(b)#

#color(green)(42)# for #color(green)(c)# gives:

#x = (-color(blue)((-19)) +- sqrt(color(blue)((-19))^2 - (4 * color(red)(-2) * color(green)(42))))/(2 * color(red)(-2))#

#x = (color(blue)(19) +- sqrt(361 - (-336)))/-4#

#x = (color(blue)(19) +- sqrt(361 + 336))/-4#

#x = (color(blue)(19) +- sqrt(697))/-4#

#x = -(color(blue)(19) +- sqrt(697))/4#