https://file.aiquickdraw.com/mathbot/file/1728219250788t8yep00p.jpg

Let's solve both parts of the problem step by step:

### Part (i): Solve \( 4x^2 - 23x + 15 = 0 \)

This is a quadratic equation, and we can use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
For \( 4x^2 - 23x + 15 = 0 \), the coefficients are:
- \( a = 4 \)
- \( b = -23 \)
- \( c = 15 \)

Now, substitute these into the quadratic formula:
\[
x = \frac{-(-23) \pm \sqrt{(-23)^2 - 4(4)(15)}}{2(4)}
\]
\[
x = \frac{23 \pm \sqrt{529 - 240}}{8}
\]
\[
x = \frac{23 \pm \sqrt{289}}{8}
\]
\[
x = \frac{23 \pm 17}{8}
\]

So, the two solutions are:
\[
x_1 = \frac{23 + 17}{8} = \frac{40}{8} = 5
\]
\[
x_2 = \frac{23 - 17}{8} = \frac{6}{8} = \frac{3}{4}
\]

Thus, the solutions for part (i) are:
\[
x = 5 \quad \text{or} \quad x = \frac{3}{4}
\]

### Part (ii): Solve \( 4(y^2 + y)^2 - 23(y^2 + y) + 15 = 0 \)

Let’s use substitution to solve this. Define:
\[
z = y^2 + y
\]
This reduces the equation to:
\[
4z^2 - 23z + 15 = 0
\]

This is another quadratic equation, and we solve it using the quadratic formula. Here, \( a = 4 \), \( b = -23 \), and \( c = 15 \) (same as in part (i)):
\[
z = \frac{-(-23) \pm \sqrt{(-23)^2 - 4(4)(15)}}{2(4)}
\]
\[
z = \frac{23 \pm \sqrt{529 - 240}}{8}
\]
\[
z = \frac{23 \pm \sqrt{289}}{8}
\]
\[
z = \frac{23 \pm 17}{8}
\]

So, the two solutions for \( z \) are:
\[
z_1 = \frac{40}{8} = 5 \quad \text{and} \quad z_2 = \frac{6}{8} = \frac{3}{4}
\]

### Now, solve for \( y \) in each case:
1. If \( z = 5 \), then:
   \[
   y^2 + y = 5
   \]
   Solve the quadratic equation:
   \[
   y^2 + y - 5 = 0
   \]
   Using the quadratic formula:
   \[
   y = \frac{-1 \pm \sqrt{1^2 - 4(1)(-5)}}{2(1)} = \frac{-1 \pm \sqrt{1 + 20}}{2} = \frac{-1 \pm \sqrt{21}}{2}
   \]
   So, \( y = \frac{-1 + \sqrt{21}}{2} \) or \( y = \frac{-1 - \sqrt{21}}{2} \).

2. If \( z = \frac{3}{4} \), then:
   \[
   y^2 + y = \frac{3}{4}
   \]
   Multiply through by 4 to clear the fraction:
   \[
   4y^2 + 4y - 3 = 0
   \]
   Solve this using the quadratic formula:
   \[
   y = \frac{-4 \pm \sqrt{4^2 - 4(4)(-3)}}{2(4)} = \frac{-4 \pm \sqrt{16 + 48}}{8} = \frac{-4 \pm \sqrt{64}}{8}
   \]
   \[
   y = \frac{-4 \pm 8}{8}
   \]
   So, \( y = \frac{-4 + 8}{8} = \frac{4}{8} = \frac{1}{2} \) or \( y = \frac{-4 - 8}{8} = \frac{-12}{8} = \frac{-3}{2} \).

### Final solutions for \( y \):
- \( y = \frac{-1 + \sqrt{21}}{2}, \frac{-1 - \sqrt{21}}{2} \)
- \( y = \frac{1}{2}, \frac{-3}{2} \)

Would you like more details on any of the steps?

### Follow-up questions:
1. How does the quadratic formula work in different types of equations?
2. What is the relationship between completing the square and the quadratic formula?
3. How can factoring be used to solve quadratic equations when possible?
4. What is the significance of the discriminant in determining the nature of the roots?
5. Can you generalize how substitution helps simplify complex equations?

**Tip**: Always check if a quadratic equation can be factored easily before resorting to the quadratic formula. This can save time and avoid unnecessary complications.

Solve 4x^2 - 23x + 15 = 0. Hence, solve 4(y^2 + y)^2 - 23(y^2 + y) + 15 = 0.

Learn how to solve the quadratic equation 4x^2 - 23x + 15 and apply substitution to solve 4(y^2 + y)^2 - 23(y^2 + y) + 15. Detailed step-by-step solution using the quadratic formula, covering key algebraic concepts. Suitable for students in Grades 8-10.

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