2x²+6x+7=0
To solve the quadratic equation \(2x^2 + 6x + 7 = 0\), you can use the quadratic formula:
\[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \]
In this equation, \(a = 2\), \(b = 6\), and \(c = 7\).
First, calculate the discriminant \(b^2 - 4ac\): \[ b^2 - 4ac = 6^2 - 4 \cdot 2 \cdot 7 = 36 - 56 = -20 \]
Since the discriminant is negative (\(-20\)), this means there are no real solutions, but there are two complex solutions.
Now plug the values into the quadratic formula: \[ x = \frac{{-6 \pm \sqrt{{-20}}}}{2 \cdot 2} = \frac{{-6 \pm \sqrt{{-20}}}}{4} \]
Simplify \(\sqrt{-20}\): \[ \sqrt{-20} = \sqrt{20}i = \sqrt{4 \cdot 5}i = 2\sqrt{5}i \]
Substitute back into the equation: \[ x = \frac{{-6 \pm 2\sqrt{5}i}}{4} = \frac{{-6}}{4} \pm \frac{{2\sqrt{5}i}}{4} = -\frac{3}{2} \pm \frac{\sqrt{5}}{2}i \]
Thus, the solutions to the equation \(2x^2 + 6x + 7 = 0\) are: \[ x = -\frac{3}{2} + \frac{\sqrt{5}}{2}i \quad \text{and} \quad x = -\frac{3}{2} - \frac{\sqrt{5}}{2}i \]
