Question 132904
Let's use the quadratic formula to solve for p:



Starting with the general quadratic


{{{ap^2+bp+c=0}}}


the general solution using the quadratic equation is:


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




So lets solve {{{p^2+4*p-2=0}}} ( notice {{{a=1}}}, {{{b=4}}}, and {{{c=-2}}})





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




{{{p = (-4 +- sqrt( 16-4*1*-2 ))/(2*1)}}} Square 4 to get 16  




{{{p = (-4 +- sqrt( 16+8 ))/(2*1)}}} Multiply {{{-4*-2*1}}} to get {{{8}}}




{{{p = (-4 +- sqrt( 24 ))/(2*1)}}} Combine like terms in the radicand (everything under the square root)




{{{p = (-4 +- 2*sqrt(6))/(2*1)}}} 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>)




{{{p = (-4 +- 2*sqrt(6))/2}}} Multiply 2 and 1 to get 2


So now the expression breaks down into two parts


{{{p = (-4 + 2*sqrt(6))/2}}} or {{{p = (-4 - 2*sqrt(6))/2}}}



Now break up the fraction



{{{p=-4/2+2*sqrt(6)/2}}} or {{{p=-4/2-2*sqrt(6)/2}}}



Simplify



{{{p=-2+sqrt(6)}}} or {{{p=-2-sqrt(6)}}}



So these expressions approximate to


{{{p=0.449489742783178}}} or {{{p=-4.44948974278318}}}



So our solutions are:

{{{p=0.449489742783178}}} or {{{p=-4.44948974278318}}}