Question 134865
{{{x^2 +6x =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^2}}} 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:


{{{(6/2)^2=3^2=9}}}


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


{{{x^2 +6x +9=2+9}}}


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


{{{(x+3)^2=11}}}


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


{{{x+3=sqrt(11)}}} or {{{x+3=-sqrt(11)}}}


Step 6:  Simplify both resulting equations:


{{{x=-3+sqrt(11)}}} or {{{x=-3-sqrt(11)}}}


And that is all there is.

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Quadratic Formula Solution:


First put your equation into standard form {{{ax^2+bx+c=0}}} by adding -2 to both sides:


{{{x^2+6x-2=0}}}


{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} where a is the coefficient on the {{{x^2}}} term, b is the coefficient on the {{{x}}} term, and c is the constant term, so:


{{{x = (-6 +- sqrt( 6^2-4*1*(-2) ))/(2*1) }}} 


{{{x = (-6 +- sqrt( 36+8 ))/2 }}}


{{{x = (-6 +- sqrt(44))/2 }}}


{{{x = (-6 +- sqrt(4*11))/2 }}}


{{{x = (-6 + 2sqrt(11))/2 }}} or {{{x = (-6 - 2sqrt(11))/2 }}}


{{{x = -3 + sqrt(11) }}} or {{{x = -3 - sqrt(11)}}}


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