Question 449322

{{{ 3n^2-2n-5=0}}}...use quadratic formula

{{{n = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} ....notice that {{{a=3}}}, {{{b=-2}}} and {{{c=-5}}}


{{{n = (-(-2) +- sqrt( (-2)^2-4*3*(-5) ))/(2*3) }}}


{{{n = (2 +- sqrt( 4+60 ))/6 }}}


{{{n = (2 +- sqrt( 64 ))/6 }}}


{{{n = (2 +- 8)/6 }}}

so, solutions are:


{{{n = (2 + 8)/6 }}}

{{{n = 10/6 }}}

{{{n = 5/3}}}

{{{n = 1.7}}}


or

{{{n = (2 - 8)/6 }}}

{{{n = -6/6 }}}

{{{n = -1}}}


let's see it on a graph:

{{{ graph( 500, 500, -10, 10, -10, 10, 3x^2-2x-5) }}}