# How do you find the discriminant and how many solutions does p^2+12p=-4 have?

May 3, 2015

A quadratic equation of the form
$a {x}^{2} + b x + c = 0$
has a discriminant $\Delta = {b}^{2} - 4 a c$

• no solutions if $D \le t a < 0$
• 1 solution if $\Delta = 0$
• 2 solutions if $\Delta > 0$

Rewriting ${p}^{2} + 12 p = - 14$
in a standard form:
${p}^{2} + 12 p + 14 = 0$

$\Delta = {\left(12\right)}^{2} - 4 \left(1\right) \left(14\right)$
$= 144 - 64$
$= 80$

Since $\Delta > 0$