Question 86296
{{{p^2}}}+5p-5=0
You'll need to apply the quadratic formula:
{{{x = (-b +- sqrt(b^2-4*a*c))/(2*a)}}} =
{{{(-5 +- sqrt(5^2-4*1*(-5)))/(2*1)}}} =
{{{(-5 +- sqrt(5^2-4*1*(-5)))/2}}} =
{{{(-5 +- sqrt(25-(-20)))/2}}} =
{{{(-5 +- sqrt(45))/2}}} =
{{{(-5 +- 3sqrt(5))/2}}}