Question 333672


{{{p^2+13p-30=0}}} Start with the given equation.



Notice that the quadratic {{{p^2+13p-30}}} is in the form of {{{Ap^2+Bp+C}}} where {{{A=1}}}, {{{B=13}}}, and {{{C=-30}}}



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



{{{p = (-B +- sqrt( B^2-4AC ))/(2A)}}} Start with the quadratic formula



{{{p = (-(13) +- sqrt( (13)^2-4(1)(-30) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=13}}}, and {{{C=-30}}}



{{{p = (-13 +- sqrt( 169-4(1)(-30) ))/(2(1))}}} Square {{{13}}} to get {{{169}}}. 



{{{p = (-13 +- sqrt( 169--120 ))/(2(1))}}} Multiply {{{4(1)(-30)}}} to get {{{-120}}}



{{{p = (-13 +- sqrt( 169+120 ))/(2(1))}}} Rewrite {{{sqrt(169--120)}}} as {{{sqrt(169+120)}}}



{{{p = (-13 +- sqrt( 289 ))/(2(1))}}} Add {{{169}}} to {{{120}}} to get {{{289}}}



{{{p = (-13 +- sqrt( 289 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{p = (-13 +- 17)/(2)}}} Take the square root of {{{289}}} to get {{{17}}}. 



{{{p = (-13 + 17)/(2)}}} or {{{p = (-13 - 17)/(2)}}} Break up the expression. 



{{{p = (4)/(2)}}} or {{{p =  (-30)/(2)}}} Combine like terms. 



{{{p = 2}}} or {{{p = -15}}} Simplify. 



So the solutions are {{{p = 2}}} or {{{p = -15}}}