How do you solve #5m ^ { 2} + 75m = - 280#?

1 Answer
Jun 6, 2017

See a solution process below:

Explanation:

First, divide each side of the equation by #color(red)(5)# which is a common factor for each term while also keeping the equation balanced:

#(5m^2 + 75m)/color(red)(5) = -280/color(red)(5)#

#(5m^2)/color(red)(5) + (75m)/color(red)(5) = -56#

#(color(red)(cancel(color(black)(5)))m^2)/cancel(color(red)(5)) + (color(red)(cancel(color(black)(75)))15m)/cancel(color(red)(5)) = -56#

#m^2 + 15m = -56#

Next, add #color(red)(56)# to each side of the equation to put this equation in standard form while keeping the equation balanced:

#m^2 + 15m + color(red)(56) = -56 + color(red)(56)#

#m^2 + 15m + 56 = 0#

We can then factor this look for two numbers which add up to 15 and multiply to 56:

#(m + 7)(m + 8) = 0#

We can now solve each term on the left side of the equation for #0#:

Solution 1)

#m + 7 = 0#

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

#m + 0 = -7#

#m = -7#

Solution 2)

#m + 8 = 0#

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

#m + 0 = -8#

#m = -8#

The solution is: #m = -7# and #m = -8#