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)#