Question 87266
Starting with the general quadratic


{{{ax^2+bx+c=0}}}


the general form of the quadratic equation is:


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


So lets solve {{{66*x^2-31*x-9=0}}}


{{{x = (31 +- sqrt( (-31)^2-4*66*-9 ))/(2*66)}}} Plug in a=66, b=-31, and c=-9




{{{x = (31 +- sqrt( 961-4*66*-9 ))/(2*66)}}} Square -31 to get 961




{{{x = (31 +- sqrt( 961+2376 ))/(2*66)}}} Multiply {{{-4*-9*66}}} to get {{{2376}}}




{{{x = (31 +- sqrt( 3337 ))/(2*66)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (31 +- sqrt(3337))/(2*66)}}} Simplify the square root




{{{x = (31 +- sqrt(3337))/132}}} Multiply 2 and 66 to get 132


So now the expression breaks down into two parts


{{{x = (31 + sqrt(3337))/132}}} or {{{x = (31 - sqrt(3337))/132}}}



Which approximate to


{{{x=0.672475548910187}}} or {{{x=-0.202778579213218}}}



So our solutions are:

{{{x=0.672475548910187}}} or {{{x=-0.202778579213218}}}


Notice when we graph {{{66*x^2-31*x-9}}} we get:


{{{ graph( 500, 500, -10.2027785792132, 10.6724755489102, -10.2027785792132, 10.6724755489102,66*x^2+-31*x+-9) }}}


when we use the root finder feature on a calculator, we find that {{{x=0.672475548910187}}} and {{{x=-0.202778579213218}}}.So this verifies our answer