How do you solve $6x^2-12x+1=0$ using the quadratic formula?

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Answer 1 · 2017-08-24T19:23:57

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:

$color(red)(6)$ for $color(red)(a)$

$color(blue)(-12)$ for $color(blue)(b)$

$color(green)(1)$ for $color(green)(c)$ gives:

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

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

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

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

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

Or

$x = 1 +- sqrt(30)/6$

How do you solve 6x^2-12x+1=06x^2-12x+1=0 using the quadratic formula?