To solve the quadratic equation \( v^2 + 9v + 20 = 0 \), we can use the quadratic formula, which states that for a quadratic equation of the form \( ax^2 + bx + c = 0 \), the solutions for \( x \) are given by:
[tex]\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
In this equation, \( a = 1 \), \( b = 9 \), and \( c = 20 \).
1. First, we need to identify the values of \( a \), \( b \), and \( c \):
[tex]\[
a = 1, \quad b = 9, \quad c = 20
\][/tex]
2. Next, we calculate the discriminant \( \Delta \), which is the part under the square root in the quadratic formula:
[tex]\[
\Delta = b^2 - 4ac
\][/tex]
Plugging in the values:
[tex]\[
\Delta = 9^2 - 4 \cdot 1 \cdot 20 = 81 - 80 = 1
\][/tex]
3. Since the discriminant (\(\Delta\)) is positive, we have two real solutions. Now we'll use the quadratic formula to find the solutions:
[tex]\[
v = \frac{-b \pm \sqrt{\Delta}}{2a}
\][/tex]
4. Substitute \( \Delta = 1 \), \( a = 1 \), and \( b = 9 \) into the formula:
[tex]\[
v = \frac{-9 \pm \sqrt{1}}{2 \cdot 1}
\][/tex]
Simplify the expression:
[tex]\[
v = \frac{-9 \pm 1}{2}
\][/tex]
5. This gives us two solutions:
- For the positive case (\( +\sqrt{1} \)):
[tex]\[
v = \frac{-9 + 1}{2} = \frac{-8}{2} = -4
\][/tex]
- For the negative case (\( -\sqrt{1} \)):
[tex]\[
v = \frac{-9 - 1}{2} = \frac{-10}{2} = -5
\][/tex]
Therefore, the solutions to the equation \( v^2 + 9v + 20 = 0 \) are:
[tex]\[
v = -4, -5
\][/tex]
