Question 98447
Well if it's not in the "equals 0" format, you must get all terms to one side



{{{6n^2= -8n-1}}} Start with the given quadratic



{{{0= -6n^2-8n-1}}} Subtract {{{6n^2}}} from both sides to get all terms to one side. Now one side equals zero




Let's use the quadratic formula to solve for n:



Starting with the general quadratic


{{{an^2+bn+c=0}}}


the general solution using the quadratic equation is:


{{{n = (-b +- sqrt( b^2-4*a*c ))/(2*a)}}}


So lets solve {{{-6*n^2-8*n-1=0}}} ( notice {{{a=-6}}}, {{{b=-8}}}, and {{{c=-1}}})


{{{n = (--8 +- sqrt( (-8)^2-4*-6*-1 ))/(2*-6)}}} Plug in a=-6, b=-8, and c=-1




{{{n = (8 +- sqrt( (-8)^2-4*-6*-1 ))/(2*-6)}}} Negate -8 to get 8




{{{n = (8 +- sqrt( 64-4*-6*-1 ))/(2*-6)}}} Square -8 to get 64  (note: remember when you square -8, you must square the negative as well. This is because {{{(-8)^2=-8*-8=64}}}.)




{{{n = (8 +- sqrt( 64+-24 ))/(2*-6)}}} Multiply {{{-4*-1*-6}}} to get {{{-24}}}




{{{n = (8 +- sqrt( 40 ))/(2*-6)}}} Combine like terms in the radicand (everything under the square root)




{{{n = (8 +- 2*sqrt(10))/(2*-6)}}} Simplify the square root (note: If you need help with simplifying the square root, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)




{{{n = (8 +- 2*sqrt(10))/-12}}} Multiply 2 and -6 to get -12


So now the expression breaks down into two parts


{{{n = (8 + 2*sqrt(10))/-12}}} or {{{n = (8 - 2*sqrt(10))/-12}}}



Now break up the fraction



{{{n=+8/-12+2*sqrt(10)/-12}}} or {{{n=+8/-12-2*sqrt(10)/-12}}}



Simplify



{{{n=-2 / 3-sqrt(10)/6}}} or {{{n=-2 / 3+sqrt(10)/6}}}



So these expressions approximate to


{{{n=-1.1937129433614}}} or {{{n=-0.139620389971937}}}



So our solutions are:

{{{n=-1.1937129433614}}} or {{{n=-0.139620389971937}}}


Notice when we graph {{{-6*x^2-8*x-1}}} (just replace n with x), we get:


{{{ graph( 500, 500, -11.1937129433614, 9.86037961002806, -11.1937129433614, 9.86037961002806,-6*x^2+-8*x+-1) }}}


when we use the root finder feature on a calculator, we find that {{{x=-1.1937129433614}}} and {{{x=-0.139620389971937}}}.So this verifies our answer