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:

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

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

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

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

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

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