Question

Solve the equation using quadratic formula.

$p^2+7p+12=0$

Short answer

The roots are$p=-4,p=-3$.

Step-by-step solution

 Given Information. 

The given equation is$p^2+7p+12=0$.

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

The quadratic equation is in standard form.

So,

$a=1,b=7,c=12$

Substitute the values in quadratic formula. 

$$\begin{gathered} p=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ p=\frac{-7\pm \sqrt{7^2-4\left(1\right)\left(12\right)}}{2\left(1\right)} \\ p=\frac{-7\pm \sqrt{49-48}}{2} \\ p=\frac{-7\pm \sqrt{1}}{2} \end{gathered}$$

Use the square root property.

role="math" localid="1648978240886" $p=\frac{-7\pm 1}{2}$

Rewrite to show two solutions. 

$$\begin{gathered} p_1=\frac{-7+1}{2} \\ {p}_1=\frac{-6}{2} \\ {p}_1=-3 \end{gathered}$$

and

$$\begin{gathered} p_2=\frac{-7-1}{2} \\ {p}_2=\frac{-8}{2} \\ {p}_2=-4 \end{gathered}$$

Thus, the solutions are$p=-4,p=-3$.