Question

Solve the equation using quadratic formula.

$p^2-6p-27=0$

Short answer

The roots are$p=-3,p=9$.

Step-by-step solution

 Given Information.    

The given equation is$p^2-6p-27=0$.

Write the quadratic equation in standard form and identify  a,b,c.

$p^2-6p-27=0$

The equation is in standard form.

So,

$a=1,b=-6,c=-27$

Substitute the values in quadratic formula.   

$$\begin{gathered} p=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ p=\frac{-(-6)\pm \sqrt{(-6{)}^2-4(1\left)\right(-27)}}{2\left(1\right)} \\ p=\frac{6\pm \sqrt{36+108}}{2} \\ p=\frac{6\pm \sqrt{144}}{2} \end{gathered}$$

Use square root property. 

$p=\frac{6\pm 12}{2}$

Rewrite to show two solutions.

$$\begin{gathered} p_1=\frac{6+12}{2} \\ {p}_1=\frac{18}{2} \\ {p}_1=9 \end{gathered}$$

and

$$\begin{gathered} p_2=\frac{6-12}{2} \\ {p}_2=\frac{-6}{2} \\ {p}_2=-3 \end{gathered}$$

Thus, the roots are$p=-3,p=9$.