SOLUTION: 9x-10 + 2x^2 + x^2+4 Triangle perimeter

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Question 131983:

9x-10 + 2x^2 + x^2+4 Triangle perimeter
Answer by checkley71(8403) About Me  (Show Source):
You can put this solution on YOUR website!
9X-10+2X^2+X^2+4
3X^2+9X-6
USING THE QUADRATIC EQUATION:x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+ WE GET:
X=(-9+-SQRT[9^2-4*3*-6])/2*3
X=(-9+-SQRT[81+72])/6
X=(-9+-SQRT153)/6
X=(-9+-12.37)/6
X=(-9+12.37)/6
X=3.37/6
X=.56 ANSWER
3*.56^2+9*.56-6
3*.3136+5.04-6
.9408+5.04-6
5.9808-6
-.0192 NOW AN ACCEPTABLE ANSWER FOR THE PERIMETER OF A TRIANGLE.