SOLUTION: please help me solve this equation x^2 +6x =2

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Question 134865This question is from textbook algebra 2
: please help me solve this equation x^2 +6x =2 This question is from textbook algebra 2

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
x%5E2+%2B6x+=2

You can either complete the square or use the quadratic formula (which is nothing more than completing the square in general terms).

Complete the square solution:

The first step is to divide both sides of the equation by the coefficient on the x%5E2 term, but since that coefficient is 1, we can skip this step.

The second step is to divide the coefficient on the x term by 2 and square the result:

%286%2F2%29%5E2=3%5E2=9

Step 3: Add the result of step 2 to both sides of the equation

x%5E2+%2B6x+%2B9=2%2B9

Step 4: You now have a perfect square trinomial on the left, so factor it:

%28x%2B3%29%5E2=11

Step 5: Take the square root of both sides of the equation, remembering to consider both positive and negative roots:

x%2B3=sqrt%2811%29 or x%2B3=-sqrt%2811%29

Step 6: Simplify both resulting equations:

x=-3%2Bsqrt%2811%29 or x=-3-sqrt%2811%29

And that is all there is.
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Quadratic Formula Solution:

First put your equation into standard form ax%5E2%2Bbx%2Bc=0 by adding -2 to both sides:

x%5E2%2B6x-2=0

x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+ where a is the coefficient on the x%5E2 term, b is the coefficient on the x term, and c is the constant term, so:

x+=+%28-6+%2B-+sqrt%28+6%5E2-4%2A1%2A%28-2%29+%29%29%2F%282%2A1%29+

x+=+%28-6+%2B-+sqrt%28+36%2B8+%29%29%2F2+

x+=+%28-6+%2B-+sqrt%2844%29%29%2F2+

x+=+%28-6+%2B-+sqrt%284%2A11%29%29%2F2+

x+=+%28-6+%2B+2sqrt%2811%29%29%2F2+ or x+=+%28-6+-+2sqrt%2811%29%29%2F2+

x+=+-3+%2B+sqrt%2811%29+ or x+=+-3+-+sqrt%2811%29

Not surprisingly (I sincerely hope), we got the same answers.