To solve the quadratic equation \( 14x^2 + 2x + 4 = 0 \), we can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 14 \), \( b = 2 \), and \( c = 4 \).
- First, calculate the discriminant \( b^2 - 4ac \):
\[ b^2 = 2^2 = 4 \] \[ 4ac = 4 \cdot 14 \cdot 4 = 224 \] \[ b^2 - 4ac = 4 - 224 = -220 \]
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Since the discriminant is negative, this indicates that there are no real solutions, but there are two complex solutions.
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Now, substitute back into the quadratic formula:
\[ x = \frac{-2 \pm \sqrt{-220}}{2 \cdot 14} \] \[ x = \frac{-2 \pm \sqrt{-220}}{28} \]
- Simplifying \( \sqrt{-220} \):
\[ \sqrt{-220} = \sqrt{220} \cdot i \] \[ \sqrt{220} = \sqrt{4 \cdot 55} = 2\sqrt{55} \] Thus, \[ \sqrt{-220} = 2\sqrt{55}i \]
- Substitute this back into our equation for \( x \):
\[ x = \frac{-2 \pm 2\sqrt{55}i}{28} \] \[ x = \frac{-1 \pm \sqrt{55}i}{14} \]
So the solutions to the equation are:
\[ x = \frac{-1 + \sqrt{55}i}{14}, \quad x = \frac{-1 - \sqrt{55}i}{14} \]
In conclusion, the solutions are complex and can be expressed as:
\[ x = \frac{-1 \pm \sqrt{55}i}{14} \]
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