Question

p^2+14p-38=0

Answer 1

use Sri Acharya's formula-----
x = (-b+- sqrt(b^2 - 4ac))/2a
for a quadratic equation of form
ax^2+bx+c=0

Now if the term b^2-4ac
>0 ---- the roots are real and distinct
=0 ----- the roots are real and equal
<0 ----- the roots are complex numbers

Answer 2

Directions say solve by completing the square. Should the problem be done differently then?

Answer 3

\[p^2+14p-38=0\]

\[p^2+14p=38\]

take half of 14 and then square
get 49
add 49 to both sides

\[p^2+14p+49=38+49\]
\[p^2+14p+49=87\]

now left hand side factors

\[(p+7)^2=87\]

take square rot

\[p+7=\pm\sqrt{87}\]

so

\[p=-7\pm\sqrt{87}\]