How do you solve #7p^2+16=2151#?

2 Answers
May 30, 2017

Answer:

#+-17.5#

Explanation:

First rearrange the equation to have all like terms on one side and unlike terms on the other:

#7p^2 = 2135#

Divide by #7# to get #p# on its own and then square root it.

#p^2 = 305#
#sqrt(p^2)= +-sqrt(305)#

#p =+-sqrt(305)#

#= +-17.4642492#

#= +-17.5 # (to 3 sf)

May 30, 2017

Answer:

See a solution process below:

Explanation:

First, subtract #color(red)(16)# from each side of the equation to isolate the #p# term while keeping the equation balanced:

#7p^2 + 16 - color(red)(16) = 2151 - color(red)(16)#

#7p^2 + 0 = 2135#

#7p^2 = 2135#

Next, divide each side of the equation by #color(red)(7)# to isolate #p^2# while keeping the equation balanced:

#(7p^2)/color(red)(7) = 2135/color(red)(7)#

#(color(red)(cancel(color(black)(7)))p^2)/cancel(color(red)(7)) = 305#

#p^2 = 305#

Now, take the square root of each side of the equation to solve for #p# while keeping the equation balanced. Remember, the square root of a number produces both a positive and negative result:

#sqrt(p^2) = +-sqrt(305)#

#p = +-17.464# rounded to the nearest thousandth.