Question 119797
{{{(1/18)h^2-(1/2)h+1=0}}} Start with the given equation




{{{(18)((1/18)h^2-(1/2)h+1)=(18)(0)}}} Multiply both sides by the LCM of 18. This will eliminate the fractions  



{{{h^2-9h+18=0}}} Distribute and multiply the LCM to each side




{{{(h-6)(h-3)=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:

{{{h-6=0}}} or  {{{h-3=0}}} 


{{{h=6}}} or  {{{h=3}}}    Now solve for h in each case



So our answer is 

 {{{h=6}}} or  {{{h=3}}} 



Notice if we graph {{{y=h^2-9h+18}}} we can see that the roots are {{{h=6}}} and  {{{h=3}}} . So this visually verifies our answer.



{{{ graph(500,500,-10,10,-10,10,0, x^2-9x+18) }}}