Question 103658


{{{2x^2-6x-3=0}}} Start with the given equation



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




Take half of the x coefficient -3 to get -1.5 (ie {{{-3/2=-1.5}}})

Now square -1.5 to get 2.25 (ie {{{(-1.5)^2=2.25}}})




{{{2(x^2-3x+2.25)=3+2.25(2)}}} Add this result (2.25) to the expression {{{x^2-3x}}}  inside the parenthesis. Now the expression {{{x^2-3x+2.25}}}  is a perfect square trinomial. Now add the result (2.25)(2) (remember we factored out a 2) to the right side.




{{{2(x-1.5)^2=3+2.25(2)}}} Factor {{{x^2-3x+2.25}}} into {{{(x-1.5)^2}}} 



{{{2(x-1.5)^2=3+4.5}}} Multiply 2.25 and 2 to get 4.5




{{{2(x-1.5)^2=7.5}}} Combine like terms on the right side


{{{(x-1.5)^2=3.75}}} Divide both sides by 2



{{{x-1.5=0+-sqrt(3.75)}}} Take the square root of both sides


{{{x=1.5+-sqrt(3.75)}}} Add 1.5 to both sides to isolate x.


So the expression breaks down to

{{{x=1.5+sqrt(3.75)}}} or {{{x=1.5-sqrt(3.75)}}}



So our answer is approximately

{{{x=3.43649167310371}}} or {{{x=-0.436491673103709}}}


Here is visual proof


{{{ graph( 500, 500, -10, 10, -10, 10, 2x^2-6x-3) }}} graph of {{{y=2x^2-6x-3}}}



When we use the root finder feature on a calculator, we would find that the x-intercepts are {{{x=3.43649167310371}}} and {{{x=-0.436491673103709}}}, so this verifies our answer.