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

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Answer 1 · 2017-12-08T18:50:32

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:

$color(red)(-12)$ for $color(red)(a)$

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

$color(green)(-12)$ for $color(green)(c)$ gives:

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

$x = (color(blue)(2) +- sqrt(color(blue)(4 - 576)))/-24$

$x = -(2 +- sqrt(-572))/24$

$x = -(2 +- sqrt(4 * -143))/24$

$x = -(2 +- sqrt(4)sqrt(-143))/24$

$x = -(2 +- 2sqrt(-143))/24$

$x = -(1 +- 1sqrt(-143))/12$

$x = -(1 +- sqrt(-143))/12$

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