Question 442552
{{{sqrt(p^2-3p+64) = p+5}}}
Square both sides.
{{{(sqrt(p^2-3p+64))^2 = (p+5)^2}}}
{{{p^2-3p+64 = p^2+10p +25}}}
{{{39 = 13p}}}
so {{{p = 3}}}

But... since we are dealing with square roots, we must check that p=3 is not an EXTRANEOUS solution.

Subsitute p = 3 in for p^2 - 3p +64

{{{3^2 - 3*3 + 64   = 64}}}

Since {{{64 >= 0}}} then {{{p =3 }}} is a valid solution.

{{{p = 3}}}