How do you solve #g^ { 2} - 14g = 51#?

2 Answers
smendyka
Mar 1, 2017

See the entire solution process below:

Explanation:

First, subtraction #color(red)(51)# from each side of the equation to create a standard form quadratic equation:

#g^2 - 14g - color(red)(51) = 51 - color(red)(51)#

#g^2 - 14g - 51 = 0#

Next, we factor the quadratic as:

#(g + 3)(g - 17) = 0#

Now, solve each term on the left side of the equation for #0#:

Solution 1)

#g + 3 = 0#

#g + 3 - color(red)(3) = 0 - color(red)(3)#

#g + 0 = -3#

#g = -3#

Solution 2)

#g - 17 = 0#

#g - 17 + color(red)(17) = 0 + color(red)(17)#

#g + 0 = 17#

#g = 17#

The solution is: #g = -3# and #g = 17#

George C.
Mar 1, 2017

#g = -3" "# or #" "g = 17#

Explanation:

Here's one method:

Given:

#g^2-14g = 51#

Add #49 = (14/2)^2# to both sides to get:

#g^2-14g+49 = 100#

Both sides of the equation are perfect squares:

#(g-7)^2 = 10^2#

We can deduce:

#g-7 = +-10#

Adding #7# to both sides we find:

#g = 7+-10#

That is:

#g = -3" "# or #" "g = 17#

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