To solve the equation \((2x - 1)(x + 3) = -2x\), we first need to expand and simplify the left-hand side and then move all terms to one side to form a standard quadratic equation.

1. Expand the left-hand side of the equation:
\[
(2x - 1)(x + 3) = 2x \cdot x + 2x \cdot 3 - 1 \cdot x - 1 \cdot 3 = 2x^2 + 6x - x - 3 = 2x^2 + 5x - 3
\]

2. Set the expanded form equal to \(-2x\):
\[
2x^2 + 5x - 3 = -2x
\]

3. Add \(2x\) to both sides to set the equation to 0:
\[
2x^2 + 5x + 2x - 3 = 0
\]
\[
2x^2 + 7x - 3 = 0
\]

Now, we have a standard quadratic equation:
\[
2x^2 + 7x - 3 = 0
\]

To solve this quadratic equation, we can use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \(a = 2\), \(b = 7\), and \(c = -3\).

4. Calculate the discriminant (\(\Delta\)):
\[
\Delta = b^2 - 4ac = 7^2 - 4 \cdot 2 \cdot (-3) = 49 + 24 = 73
\]

5. Substitute the values into the quadratic formula:
\[
x = \frac{-7 \pm \sqrt{73}}{2 \cdot 2}
\]
\[
x = \frac{-7 \pm \sqrt{73}}{4}
\]

Thus, the solutions are:
\[
x = \frac{-7 + \sqrt{73}}{4} \quad \text{and} \quad x = \frac{-7 - \sqrt{73}}{4}
\]

Would you like more details or have any questions about this solution?

Here are some related questions for further exploration:

1. How do you factor quadratic equations?
2. How can you complete the square to solve quadratic equations?
3. What are the different methods to solve quadratic equations?
4. How do you determine the vertex of a quadratic equation?
5. What is the significance of the discriminant in quadratic equations?
6. How can you graph a quadratic equation?
7. What are real-world applications of quadratic equations?
8. How can you solve systems of equations involving a quadratic and a linear equation?

**Tip:** Always check your solutions by substituting them back into the original equation to verify they satisfy the equation.

Learn how to solve the quadratic equation (2x-1)(x+3)=-2x step-by-step with a detailed solution. This problem involves expanding and simplifying polynomial equations, applying the quadratic formula, and understanding algebraic concepts. Suitable for high school students.

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