b.  x2+13x+40=0𝑥2+13𝑥+40=0 Type the answer in the box below.x=𝑥=   or   x=𝑥=  Show your reasoning.

1. Break Down the Problem

The given equation is a quadratic equation: $x^2 + 13x + 40 = 0$ To solve it, we will use the quadratic formula: $x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}$ where $a = 1$, $b = 13$, and $c = 40$.

2. Relevant Concepts

Using the quadratic formula, we need to calculate the discriminant first: $D = b^2 - 4ac$

3. Analysis and Detail

1. Substitute the values of $a$, $b$, and $c$: $D = 13^2 - 4 \cdot 1 \cdot 40$ $D = 169 - 160 = 9$

2. Since the discriminant $D$ is positive, there are two distinct real solutions. Now apply the quadratic formula: $x = \frac{{-13 \pm \sqrt{9}}}{2 \cdot 1}$ $x = \frac{{-13 \pm 3}}{2}$

3. Calculate the two possible values for $x$:

   • For the positive case: $x_1 = \frac{{-13 + 3}}{2} = \frac{{-10}}{2} = -5$
   • For the negative case: $x_2 = \frac{{-13 - 3}}{2} = \frac{{-16}}{2} = -8$

4. Verify and Summarize

The two solutions for the quadratic equation are $x = -5$ and $x = -8$. We can verify by substituting back into the original equation:

• For $x = -5$: $(-5)^2 + 13(-5) + 40 = 25 - 65 + 40 = 0$
• For $x = -8$: $(-8)^2 + 13(-8) + 40 = 64 - 104 + 40 = 0$ Both values satisfy the equation.

Final Answer

The solutions to the equation $x^2 + 13x + 40 = 0$ are: $x = -5 \quad \text{or} \quad x = -8$