How do you solve #-g ^ { 2} + 6g - 2= 0#?

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Answer 1 · 2017-08-28T13:40:20

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)(6)# for #color(blue)(b)#

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

#g = (-color(blue)(6) +- sqrt(color(blue)(6)^2 - (4 * color(red)(-1) * color(green)(-2))))/(2 * color(red)(-1))#

#g = (-color(blue)(6) +- sqrt(36 - 8))/-2#

#g = (-color(blue)(6) +- sqrt(28))/-2#

#g = (-color(blue)(6) +- sqrt(4 * 7))/-2#

#g = (-color(blue)(6) +- sqrt(4)sqrt(7))/-2#

#g = (-color(blue)(6) +- 2sqrt(7))/-2#

#g = 3 +- sqrt(7)#