SOLUTION: find the real solutions to this equation:(x+3)^2+9(x+3)+14=0

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Question 339150: find the real solutions to this equation:(x+3)^2+9(x+3)+14=0
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
%28x%2B3%29%5E2%2B9%28x%2B3%29%2B14=0 Start with the given equation.


x%5E2%2B6x%2B9%2B9%28x%2B3%29%2B14=0 FOIL


x%5E2%2B6x%2B9%2B9x%2B27%2B14=0 Distribute.


x%5E2%2B15x%2B50=0 Combine like terms.


Notice that the quadratic x%5E2%2B15x%2B50 is in the form of Ax%5E2%2BBx%2BC where A=1, B=15, and C=50


Let's use the quadratic formula to solve for "x":


x+=+%28-B+%2B-+sqrt%28+B%5E2-4AC+%29%29%2F%282A%29 Start with the quadratic formula


x+=+%28-%2815%29+%2B-+sqrt%28+%2815%29%5E2-4%281%29%2850%29+%29%29%2F%282%281%29%29 Plug in A=1, B=15, and C=50


x+=+%28-15+%2B-+sqrt%28+225-4%281%29%2850%29+%29%29%2F%282%281%29%29 Square 15 to get 225.


x+=+%28-15+%2B-+sqrt%28+225-200+%29%29%2F%282%281%29%29 Multiply 4%281%29%2850%29 to get 200


x+=+%28-15+%2B-+sqrt%28+25+%29%29%2F%282%281%29%29 Subtract 200 from 225 to get 25


x+=+%28-15+%2B-+sqrt%28+25+%29%29%2F%282%29 Multiply 2 and 1 to get 2.


x+=+%28-15+%2B-+5%29%2F%282%29 Take the square root of 25 to get 5.


x+=+%28-15+%2B+5%29%2F%282%29 or x+=+%28-15+-+5%29%2F%282%29 Break up the expression.


x+=+%28-10%29%2F%282%29 or x+=++%28-20%29%2F%282%29 Combine like terms.


x+=+-5 or x+=+-10 Simplify.


So the solutions are x+=+-5 or x+=+-10


If you need more help, email me at jim_thompson5910@hotmail.com

Also, feel free to check out my tutoring website.

Jim