How do you find the zeros, real and imaginary, of $y= x^2+12x+6$ using the quadratic formula?

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Answer 1 · 2017-08-12T21:46:29

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 $color(blue)(b)$

$color(green)(6)$ for $color(green)(c)$ gives:

$x = (-color(blue)(12) +- sqrt(color(blue)(12)^2 - (4 * color(red)(1) * color(green)(6))))/(2 * color(red)(1))$

$x = (-color(blue)(12) +- sqrt(144 - 24))/2$

$x = (-color(blue)(12) +- sqrt(120))/2$

$x = (-color(blue)(12) +- sqrt(4 * 30))/2$

$x = (-color(blue)(12) - sqrt(4 * 30))/2$ and $x = (-color(blue)(12) + sqrt(4 * 30))/2$

$x = (-color(blue)(12) - sqrt(4)sqrt(30))/2$ and $x = (-color(blue)(12) + sqrt(4)sqrt(30))/2$

$x = (-color(blue)(12) - 2sqrt(30))/2$ and $x = (-color(blue)(12) + 2sqrt(30))/2$

$x = -color(blue)(12)/2 - (2sqrt(30))/2$ and $x = -color(blue)(12)/2 + (2sqrt(30))/2$

$x = -6 - sqrt(30)$ and $x = -6 + sqrt(30)$

How do you find the zeros, real and imaginary, of y= x^2+12x+6 y= x^2+12x+6 using the quadratic formula?