How do you solve #g^ { 2} - 2g - 47= 0#?

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Answer 1 · 2017-10-08T17:30:12

See a solution process below:

We can use the quadratic equation to solve this problem:

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)(-2)# for #color(blue)(b)#

#color(green)(-47)# for #color(green)(c)# gives:

#g = (-color(blue)((-2)) +- sqrt(color(blue)((-2))^2 - (4 * color(red)(1) * color(green)(-47))))/(2 * color(red)(1))#

#g = (2 +- sqrt(4 - (-188)))/2#

#g = (2 +- sqrt(4 + 188))/2#

#g = (2 +- sqrt(192))/2#

#g = (2 - sqrt(64 * 3))/2# and #g = (2 + sqrt(64 * 3))/2#

#g = (2 - sqrt(64)sqrt(3))/2# and #g = (2 + sqrt(64)sqrt(3))/2#

#g = (2 - 8sqrt(3))/2# and #g = (2 + 8sqrt(3))/2#

#g = 2/2 - (8sqrt(3))/2# and #g = 2/2 + (8sqrt(3))/2#

#g = 1 - 4sqrt(3)# and #g = 1 + 4sqrt(3)#