In the following questions two equation numbered I and II are given You have to solve both equations and . . . . .
(i) `x^(2)+x-20=0`
(ii) `y^(2)-y-30=0`

Answer

To solve the given equations, we will follow the steps outlined below:

### Step 1: Solve the first equation \(x^2 + x - 20 = 0\)

1. **Identify the quadratic equation**: The equation is in the standard form \(ax^2 + bx + c = 0\), where \(a = 1\), \(b = 1\), and \(c = -20\).

2. **Factor the equation**: We need to find two numbers that multiply to \(-20\) (the constant term) and add up to \(1\) (the coefficient of \(x\)). The numbers \(5\) and \(-4\) satisfy this condition because:
- \(5 \times (-4) = -20\)
- \(5 + (-4) = 1\)

3. **Rewrite the equation**: We can rewrite the equation as:
\[ x^2 + 5x - 4x - 20 = 0 \]

4. **Group the terms**: Group the terms to factor by grouping:
\[ (x^2 + 5x) + (-4x - 20) = 0 \]

5. **Factor out common terms**:
\[ x(x + 5) - 4(x + 5) = 0 \]
This can be factored as:
\[ (x + 5)(x - 4) = 0 \]

6. **Find the solutions**: Set each factor to zero:
\[ x + 5 = 0 \quad \Rightarrow \quad x = -5 \]
\[ x - 4 = 0 \quad \Rightarrow \quad x = 4 \]

### Step 2: Solve the second equation \(y^2 - y - 30 = 0\)

1. **Identify the quadratic equation**: The equation is in the standard form \(ay^2 + by + c = 0\), where \(a = 1\), \(b = -1\), and \(c = -30\).

2. **Factor the equation**: We need to find two numbers that multiply to \(-30\) and add up to \(-1\). The numbers \(6\) and \(-5\) satisfy this condition because:
- \(6 \times (-5) = -30\)
- \(6 + (-5) = -1\)

3. **Rewrite the equation**: We can rewrite the equation as:
\[ y^2 - 6y + 5y - 30 = 0 \]

4. **Group the terms**: Group the terms to factor by grouping:
\[ (y^2 - 6y) + (5y - 30) = 0 \]

5. **Factor out common terms**:
\[ y(y - 6) + 5(y - 6) = 0 \]
This can be factored as:
\[ (y - 6)(y + 5) = 0 \]

6. **Find the solutions**: Set each factor to zero:
\[ y - 6 = 0 \quad \Rightarrow \quad y = 6 \]
\[ y + 5 = 0 \quad \Rightarrow \quad y = -5 \]

### Final Solutions
- For the first equation \(x^2 + x - 20 = 0\), the solutions are \(x = -5\) and \(x = 4\).
- For the second equation \(y^2 - y - 30 = 0\), the solutions are \(y = 6\) and \(y = -5\).