https://file.aiquickdraw.com/mathbot/file/172649747138225oz3z1q.jpg

The image contains six quadratic equations that need to be solved using the quadratic formula. Here's a step-by-step breakdown of each equation:

### Quadratic Formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Where \(a\), \(b\), and \(c\) are coefficients from the general quadratic equation \(ax^2 + bx + c = 0\).

### Solutions:

1. **Equation (a):** \(x^2 + x - 1 = 0\)

   - Coefficients: \(a = 1\), \(b = 1\), \(c = -1\)
   - Using the quadratic formula:
     \[
     x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-1)}}{2(1)} = \frac{-1 \pm \sqrt{1 + 4}}{2} = \frac{-1 \pm \sqrt{5}}{2}
     \]
     Solutions:
     \[
     x = \frac{-1 + \sqrt{5}}{2}, \quad x = \frac{-1 - \sqrt{5}}{2}
     \]

2. **Equation (b):** \(2x^2 = 5x + 3\)

   Rearranging:
   \[
   2x^2 - 5x - 3 = 0
   \]
   - Coefficients: \(a = 2\), \(b = -5\), \(c = -3\)
   - Applying the quadratic formula:
     \[
     x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(2)(-3)}}{2(2)} = \frac{5 \pm \sqrt{25 + 24}}{4} = \frac{5 \pm \sqrt{49}}{4}
     \]
     Solutions:
     \[
     x = \frac{5 + 7}{4} = 3, \quad x = \frac{5 - 7}{4} = -\frac{1}{2}
     \]

3. **Equation (c):** \(2x^2 + x - 5 = 0\)

   - Coefficients: \(a = 2\), \(b = 1\), \(c = -5\)
   - Applying the quadratic formula:
     \[
     x = \frac{-1 \pm \sqrt{1^2 - 4(2)(-5)}}{2(2)} = \frac{-1 \pm \sqrt{1 + 40}}{4} = \frac{-1 \pm \sqrt{41}}{4}
     \]
     Solutions:
     \[
     x = \frac{-1 + \sqrt{41}}{4}, \quad x = \frac{-1 - \sqrt{41}}{4}
     \]

4. **Equation (d):** \(x^2 + 7x - 1 = 0\)

   - Coefficients: \(a = 1\), \(b = 7\), \(c = -1\)
   - Applying the quadratic formula:
     \[
     x = \frac{-7 \pm \sqrt{7^2 - 4(1)(-1)}}{2(1)} = \frac{-7 \pm \sqrt{49 + 4}}{2} = \frac{-7 \pm \sqrt{53}}{2}
     \]
     Solutions:
     \[
     x = \frac{-7 + \sqrt{53}}{2}, \quad x = \frac{-7 - \sqrt{53}}{2}
     \]

5. **Equation (e):** \((x+1)(x-5) = -3\)

   Expanding:
   \[
   x^2 - 5x + x - 5 = -3 \quad \Rightarrow \quad x^2 - 4x - 5 = -3
   \]
   Rearranging:
   \[
   x^2 - 4x - 2 = 0
   \]
   - Coefficients: \(a = 1\), \(b = -4\), \(c = -2\)
   - Applying the quadratic formula:
     \[
     x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(-2)}}{2(1)} = \frac{4 \pm \sqrt{16 + 8}}{2} = \frac{4 \pm \sqrt{24}}{2} = \frac{4 \pm 2\sqrt{6}}{2}
     \]
     Solutions:
     \[
     x = 2 + \sqrt{6}, \quad x = 2 - \sqrt{6}
     \]

6. **Equation (f):** \(3x^2 = 4x\)

   Rearranging:
   \[
   3x^2 - 4x = 0 \quad \Rightarrow \quad x(3x - 4) = 0
   \]
   Solutions:
   \[
   x = 0, \quad 3x - 4 = 0 \quad \Rightarrow \quad x = \frac{4}{3}
   \]

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### Related Questions:
1. How do you derive the quadratic formula?
2. What are the discriminant's roles in quadratic equations?
3. How do you solve quadratic equations by factoring?
4. Can the quadratic formula give complex solutions?
5. When is completing the square useful in solving quadratic equations?

### Tip:
Always check the discriminant (\(b^2 - 4ac\)) before solving to determine the nature of the roots (real or complex).

Solve the following quadratic equations using the quadratic formula.

Learn how to solve quadratic equations such as x^2 + x - 1 = 0 and 2x^2 - 5x - 3 = 0 using the quadratic formula. This step-by-step solution is ideal for students in Grades 9-11 and covers essential algebra concepts.

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