Question 113450
{{{3x^2-18x+ 24 = 18}}} Start with the given equation



{{{3x^2-18x=-6}}} Subtract 24 from both sides


{{{3(x^2-6x)=-6}}} Factor out the leading coefficient 3.  This step is important since we want the {{{x^2}}} coefficient to be equal to 1.




Take half of the x coefficient -6 to get -3 (ie {{{-6/2=-3}}})

Now square -3 to get 9 (ie {{{(-3)^2=9}}})




{{{3(x^2-6x+9)=-6+9(3)}}} Add this result (9) to the expression {{{x^2-6x}}}  inside the parenthesis. Now the expression {{{x^2-6x+9}}}  is a perfect square trinomial. Now add the result (9)(3) (remember we factored out a 3) to the right side.




{{{3(x-3)^2=-6+9(3)}}} Factor {{{x^2-6x+9}}} into {{{(x-3)^2}}}  (note: if you need help with factoring, check out this <a href=http://www.algebra.com/algebra/homework/playground/change-this-name4450.solver>solver</a>)



{{{3(x-3)^2=-6+27}}} Multiply 9 and 3 to get 27




{{{3(x-3)^2=21}}} Combine like terms on the right side


{{{(x-3)^2=7}}} Divide both sides by 3



{{{x-3=0+-sqrt(7)}}} Take the square root of both sides


{{{x=3+-sqrt(7)}}} Add 3 to both sides to isolate x.


So the expression breaks down to

{{{x=3+sqrt(7)}}} or {{{x=3-sqrt(7)}}}



So our answer is approximately

{{{x=5.64575131106459}}} or {{{x=0.354248688935409}}}