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:

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

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

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

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

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

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

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

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

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