Question 87558
{{{66x^2 + 67x + 10 =0}}}  This one looks intimidating, when you apply the quadratic formula.  Actually, it's not so bad, once you get past the {{{b^2-4ac}}}!


{{{66x^2 + 67x + 10 =0}}}
a=66, b=67, c=10


{{{x= (-b+-sqrt(b^2-4ac))/(2a)}}}

{{{x= (-67+-sqrt(67^2-4*66*10))/(2*66)}}}

{{{x= (-67+-sqrt(1849))/(132)}}}

{{{x= (-67+-43)/(132)}}}
{{{x=-24/132}}}  or {{{x=-110/132}}}
{{{x=-2/11}}}  or {{{x= -55/66=-5/6}}}


Of course, this means that the problem could have been solved by factoring!!

{{{(11x+2)(6x+5)=0 }}}


Another really interesting problem!!


R^2 Retired from SCC