Question 547274
{{{2p^(2)-119=3p}}}

Since we have a {{{p^2}}} term (and this is the biggest term), a {{{p}}} term and a constant, it is a quadratic.  We can put it into {{{ax^2+bx+c=0}}} form. 

 {{{2p^(2)-3p-119=0}}}

We could try to think of two numbers that multiply to {{{2*-119}}} and add to {{{-3}}}.  But this may be hard to think of.  Instead we will use the quadratic formula which states (I will use p instead of x, though x is most commonly used):
{{{p=(-(b)+ sqrt((b)^2-4(a)(c)))/2(a)}}}
or 
{{{p=(-(b)- sqrt((b)^2-4(a)(c)))/2(a)}}}



{{{p=(-(-3)+ sqrt((-3)^2-4(2)(-119)))/2(2)}}}
{{{p=(3+ sqrt(9+952))/4}}}
{{{p=(3+ sqrt(961))/4}}}
{{{p=(3+ 31)/4}}}
{{{p=(3+31)/4}}}
{{{p=34/4}}}
{{{p=8.5}}}
OR
{{{p=(-(-3)- sqrt((-3)^2-4(2)(-119)))/2(2)}}}
{{{p=(3- sqrt(9+952))/4}}}
{{{p=(3- sqrt(961))/4}}}
{{{p=(3- 31)/4}}}
{{{p=(3-31)/4}}}
{{{p=-28/4}}}
{{{p=-7}}}