Question 131486
{{{3x^2=2x+5}}}


The first step is to put the equation into standard form ({{{ax^2+bx+c=0}}}, so:


Add -2x-5 to both sides:


{{{3x^2-2x-5=0}}}


There are three methods to solve this.  The first is by factoring the trinomial into to binomials:


{{{(3x-5)(x+1)=0}}}, therefore {{{3x-5=0}}} or {{{x+1=0}}}, and {{{x=5/3}}} or {{{x=-1}}}


The second method is to Complete the Square:


1.  Put the constant term back on the right:
{{{3x^2-2x=5}}}


2.  Divide through by the lead coefficient:
{{{x^2-2x/3=5/3}}}


3.  Divide the coefficient on the x term by 2, square it, and add the result to both sides.  {{{(-2/3)/2=-1/3}}}, {{{(-1/3)^2=1/9}}}, so:
{{{x^2-2x/3+1/9=5/3+1/9}}}


4.  Simplify the right:
{{{x^2-2x/3+1/9=16/9}}}


5.  Now that the left is a perfect square, factor it:
{{{(x-(1/3))^2=16/9}}}


6.  Take the square root of both sides:
{{{x-(1/3)=sqrt(16/9)}}} or {{{x-(1/3)=-sqrt(16/9)}}}
{{{x-(1/3)=4/3}}} or {{{x-(1/3)=-4/3}}}
{{{x=5/3}}} or {{{x=-3/3=-1}}}


The third way is to use the quadratic formula:
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}.  For your problem, a = 3, b = -2, and c = -5, so:


{{{x = (-(-2) +- sqrt( (-2)^2-4*3*(-5)))/(2*3) }}} 
{{{x = (2 +- sqrt( 4-(-60)))/6 }}} 
{{{x = (2 +- sqrt(64))/6 }}}
{{{x = (2 + 8)/6 }}} or  {{{x = (2 - 8)/6 }}}
{{{x = 10/6 =5/3}}} or  {{{x = -6/6 =-1}}} 


Which way seems easiest to you?  You tell me.  The factoring method certainly looks easy, but in truth, I didn't see the factorization until after I had used the quadratic formula and found that there were rational roots to the equation.  So the quadratic formula was really the easiest for me.  I generally avoid completing the square when dividing by the lead coefficient or dividing the first degree term coefficient by 2 results in a fraction because it can become a messy calculation.  The fact that this problem had rational roots made it all come out very neatly, but neat perfect squares are not something that you will generally find in real life applications.