Question 120407


{{{6x^2+x-12=0}}} Start with the given equation


{{{(2x+3)(3x-4)=0}}} Factor the left side (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>)




Now set each factor equal to zero:

{{{2x+3=0}}} or  {{{3x-4=0}}} 


{{{x=-3/2}}} or  {{{x=4/3}}}    Now solve for x in each case



So our answer is 

 {{{x=-3/2}}} or  {{{x=4/3}}} 



Notice if we graph {{{y=6x^2+x-12}}}  we can see that the roots are {{{x=-3/2}}} and  {{{x=4/3}}} . So this visually verifies our answer.



{{{ graph(500,500,-10,10,-10,10,0, 6x^2+x-12) }}}