How do you solve #g^2 + g = 56#?

1 Answer
Jan 18, 2018

See a solution process below if you are trying to solve for #g#:

Explanation:

First, put the equation in standard form:

#g^2 + g = 56#

#g^2 + g - color(red)(56) = 56 - color(red)(56)#

#g^2 + g - 56 = 0#

Next, factor the left side of the equation as:

#(g + 8)(g - 7) = 0#

Now, solve each term on the left for #0# to find the solutions:

Solution 1:

#g + 8 = 0#

#g + 8 - color(red)(8) = 0 - color(red)(8)#

#g + 0 = -8#

#g = -8#

Solution 2:

#g - 7 = 0#

#g - 7 + color(red)(7) = 0 + color(red)(7)#

#g - 0 = 7#

#g = 7#

The Solution Is:

#g = {-8, 7}#

Another process is to use the quadratic formula:

The quadratic formula states:

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)(-56)# for #color(green)(c)# gives:

#g = (-color(blue)(1) +- sqrt(color(blue)(1)^2 - (4 * color(red)(1) * color(green)(-56))))/(2 * color(red)(1))#

#g = (-1 +- sqrt(1 - (-224)))/2#

#g = (-1 +- sqrt(1 + 224))/2#

#g = (-1 +- sqrt(225))/2#

#g = (-1 - sqrt(225))/2# or #g = (-1 + sqrt(225))/2#

#g = (-1 - 15)/2# or #g = (-1 + 15)/2#

#g = (-16)/2# or #g = 14/2#

#g = -8# or #g = 7#

#g = {-8, 7}#