Question 191085
First, let's complete the square for the left side of the equation {{{3x^2+12x-9=0}}}




{{{3x^2+12x-9}}} Start with the given expression.



{{{3(x^2+4x-3)}}} Factor out the {{{x^2}}} coefficient {{{3}}}. This step is very important: the {{{x^2}}} coefficient <font size=4><b>must</b></font> be equal to 1.



Take half of the {{{x}}} coefficient {{{4}}} to get {{{2}}}. In other words, {{{(1/2)(4)=2}}}.



Now square {{{2}}} to get {{{4}}}. In other words, {{{(2)^2=(2)(2)=4}}}



{{{3(x^2+4x+highlight(4-4)-3)}}} Now add <font size=4><b>and</b></font> subtract {{{4}}} inside the parenthesis. Make sure to place this after the "x" term. Notice how {{{4-4=0}}}. So the expression is not changed.



{{{3((x^2+4x+4)-4-3)}}} Group the first three terms.



{{{3((x+2)^2-4-3)}}} Factor {{{x^2+4x+4}}} to get {{{(x+2)^2}}}.



{{{3((x+2)^2-7)}}} Combine like terms.



{{{3(x+2)^2+3(-7)}}} Distribute.



{{{3(x+2)^2-21}}} Multiply.



So after completing the square, {{{3x^2+12x-9}}} transforms to {{{3(x+2)^2-21}}}. So {{{3x^2+12x-9=3(x+2)^2-21}}}.



So {{{3x^2+12x-9=0}}} is equivalent to {{{3(x+2)^2-21=0}}}.



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Now let's solve {{{3(x+2)^2-21=0}}}



{{{3(x+2)^2-21=0}}} Start with the given equation.



{{{3(x+2)^2=21}}} Add 21 to both sides



{{{(x+2)^2=(21)/(3)}}} Divide both sides by {{{3}}}.



{{{(x+2)^2=7}}} Reduce.



{{{x+2=""+-sqrt(7)}}} Take the square root of both sides. (note: don't forget the "plus/minus")



{{{x+2=sqrt(7)}}} or {{{x+2=-sqrt(7)}}} Break up the "plus/minus" to form two equations.



{{{x=-2+sqrt(7)}}} or {{{x=-2-sqrt(7)}}} Subtract {{{2}}} from both sides.



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Answer:



So the solutions are {{{x=-2+sqrt(7)}}} or {{{x=-2-sqrt(7)}}}.