Questions: 2p^2 + p + 13 = 8

Solution Steps

We start with the equation given in the problem:

\[ 2p^{2} + p + 13 = 8 \]

To solve for \( p \), we rearrange the equation by moving all terms to one side:

\[ 2p^{2} + p + 13 - 8 = 0 \]

This simplifies to:

\[ 2p^{2} + p + 5 = 0 \]

Next, we attempt to factor the polynomial \( 2p^{2} + p + 5 \). However, it is determined that this polynomial does not factor nicely over the integers, and thus remains as:

\[ 2p^{2} + p + 5 \]

To find the solutions for \( p \), we apply the quadratic formula. The solutions to the equation \( 2p^{2} + p + 5 = 0 \) are given by:

\[ p = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} \]

where \( a = 2 \), \( b = 1 \), and \( c = 5 \). Substituting these values into the formula yields:

\[ p = \frac{-1 \pm \sqrt{1^{2} - 4 \cdot 2 \cdot 5}}{2 \cdot 2} \]

Calculating the discriminant:

\[ 1 - 40 = -39 \]

Thus, the solutions become:

\[ p = \frac{-1 \pm \sqrt{-39}}{4} \]

This can be expressed as:

\[ p = \frac{-1}{4} \pm \frac{\sqrt{39}i}{4} \]

The final solutions for \( p \) are:

\[ p = -\frac{1}{4} - \frac{\sqrt{39}i}{4} \quad \text{and} \quad p = -\frac{1}{4} + \frac{\sqrt{39}i}{4} \]

Final Answer