How do you use the quadratic formula to solve #x^2+4x-2=0#?

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Answer 1 · 2018-06-12T18:39:22

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)(4)# for #color(blue)(b)#

#color(green)(-2)# for #color(green)(c)# gives:

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

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

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

#x = (-color(blue)(4) +- sqrt(4 * 6))/2#

#x = (-color(blue)(4) +- (sqrt(4)sqrt(6)))/2#

#x = (-color(blue)(4) +- 2sqrt(6))/2#

#x = -color(blue)(4)/2 +- (2sqrt(6))/2#

#x = -2 +- sqrt(6)#

#x = {(-2 - sqrt(6)), (-2 + sqrt(6))}#