Question 849610
First, you need to know that the perimeter of a triangle is 

P = a + b + c.
 

Since we already know the perimeter, we simply plug that into the formula along with the three sides given and we simply solve for x.

18 = (x^2-x) + (2x+1) + (x+2).

Combine like terms and simplify.

18 = x^2-x + 3x + 3

18 = x^2+2x + 3. Since we have a second degree equation, this is quadratic. Therefore we will need to make the entire equation equal to zero to find the values of x, or the roots of the equation. Do this by subtracting 18 from both sides.

x^2+2x-15 = 0. Now use the quadratic formula to find x:


*[invoke quadratic "x", 1, 2, -15 ]

From this you can see that x is either -5 or 3. 

Let's plug in each value of x into the original equation to see which answer we can disregard and which one we should accept as the correct value for x.

18 = (x^2-x) + (2x+1) + (x+2)


18 = (-5^2-5) + (2(-5)+1) + (-5+2)

18 = 20 -9 + -3 

18 = 11-3

18 = 8. This is obviously false so we can disregard -5 since it is an extraneous root. Now let's try the equation when x = 3.


18 = (3^2-3) + (2(3)+1) + (3+2)

18 = 6 + 7 + 5

18 = 18 This checks out. Therefore the lengths of the different sides of the triangle are 5,6,and 7.