Questions: Solve the following equation. v^2 + 3v - 40 = 0

Solution Steps

To solve the quadratic equation \( v^2 + 3v - 40 = 0 \), we can use the quadratic formula, which is given by:

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

where \( a \), \( b \), and \( c \) are the coefficients from the equation \( av^2 + bv + c = 0 \). In this case, \( a = 1 \), \( b = 3 \), and \( c = -40 \). We will calculate the discriminant \( b^2 - 4ac \) to determine the solutions.

The given quadratic equation is

\[ v^2 + 3v - 40 = 0 \]

From this equation, we identify the coefficients as follows:

• \( a = 1 \)
• \( b = 3 \)
• \( c = -40 \)

We calculate the discriminant \( D \) using the formula

\[ D = b^2 - 4ac \]

Substituting the values of \( a \), \( b \), and \( c \):

\[ D = 3^2 - 4 \cdot 1 \cdot (-40) = 9 + 160 = 169 \]

Since the discriminant is positive, we can find two real solutions using the quadratic formula:

\[ v = \frac{{-b \pm \sqrt{D}}}{2a} \]

Calculating the two solutions:

1. For \( v_1 \):

\[ v_1 = \frac{{-3 + \sqrt{169}}}{2 \cdot 1} = \frac{{-3 + 13}}{2} = \frac{10}{2} = 5.0 \]

2. For \( v_2 \):

\[ v_2 = \frac{{-3 - \sqrt{169}}}{2 \cdot 1} = \frac{{-3 - 13}}{2} = \frac{-16}{2} = -8.0 \]

Final Answer