Question 449247
{{{3y^2 + 14y = 5}}}....use quadratic formula to solve for {{{y}}}

{{{3y^2 + 14y - 5 = 0}}}

{{{y = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}....{{{a=3}}}, {{{b=14}}} and {{{c=-5}}}


{{{y= (-14 +- sqrt(14^2-4*3*(-5) ))/(2*3) }}}


{{{y= (-14 +- sqrt(196+60))/6 }}}


{{{y= (-14 +- sqrt(256))/6 }}}


{{{y= (-14 +- 16)/6 }}}

so, solutions are:

{{{y= (-14 + 16)/6 }}}


{{{y= 2/6 }}}


{{{y= 13}}}

or


{{{y= (-14 - 16)/6 }}}


{{{y= -30/6 }}}


{{{y= -5}}}


let's see it on a graph:


{{{ graph( 500, 500, -10,10, -25, 10, 3x^2 + 14x - 5 ) }}}