How do you solve $x^2+4x+2=0$ using the quadratic formula?

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Answer 1 · 2018-03-04T23:31:53

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:

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

For: $1x^2 + 4x + 2 = 0$

We can substitute:

$color(red)(1)$ for $color(red)(a)$

$color(blue)(4)$ for $color(blue)(b)$

$color(green)(2)$ for $color(green)(c)$ giving:

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

$x = (-color(blue)(4) +- sqrt(16 - 8))/2$

$x = (-color(blue)(4) - sqrt(8))/2$ and $x = (-color(blue)(4) + sqrt(8))/2$

$x = -color(blue)(4)/2 - sqrt(8)/2$ and $x = -color(blue)(4)/2 + sqrt(8)/2$

$x = -2 - sqrt(8)/2$ and $x = -2 + sqrt(8)/2$

$x = -2 - sqrt(4 xx 2)/2$ and $x = -2 + sqrt(4 xx 2)/2$

$x = -2 - (sqrt(4)sqrt(2))/2$ and $x = -2 + (sqrt(4)sqrt(2))/2$

$x = -2 - (2sqrt(2))/2$ and $x = -2 + (2sqrt(2))/2$

$x = -2 - sqrt(2)$ and $x = -2 + sqrt(2)$

How do you solve x^2+4x+2=0x^2+4x+2=0 using the quadratic formula?