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

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Answer 1 · 2017-08-24T18:16:40

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)(1)# for #color(blue)(b)#

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

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

#x = (-color(blue)(1) +- sqrt(1 - (-4)))/(-2)#

#x = (-color(blue)(1) +- sqrt(1 + 4))/(-2)#

#x = (-color(blue)(1) +- sqrt(5))/(-2)#

#x = (color(blue)(1) +- sqrt(5))/(2)#

Or

#1/2 +- sqrt(5)/2#