Solve :
(i) `2x^(2)-5x+3=0`
(ii) `(x^(2)+3x)^(2)-(x^(2)+3x)-6=0, xepsilonR`

Answer

Let's solve the given quadratic equations step by step.

### (i) Solve: \( 2x^2 - 5x + 3 = 0 \)

**Step 1: Identify the coefficients.**
The given quadratic equation is in the standard form \( ax^2 + bx + c = 0 \).
Here, \( a = 2 \), \( b = -5 \), and \( c = 3 \).

**Step 2: Factor the quadratic equation.**
We need to find two numbers that multiply to \( ac = 2 \times 3 = 6 \) and add to \( b = -5 \). The numbers are -2 and -3.

**Step 3: Rewrite the equation using these numbers.**
We can split the middle term:
\[ 2x^2 - 2x - 3x + 3 = 0 \]

**Step 4: Group the terms.**
Group the first two and the last two terms:
\[ (2x^2 - 2x) + (-3x + 3) = 0 \]

**Step 5: Factor by grouping.**
Factor out the common terms:
\[ 2x(x - 1) - 3(x - 1) = 0 \]
Now, factor out \( (x - 1) \):
\[ (2x - 3)(x - 1) = 0 \]

**Step 6: Set each factor to zero.**
Now we can set each factor to zero:
1. \( 2x - 3 = 0 \)
2. \( x - 1 = 0 \)

**Step 7: Solve for \( x \).**
From \( 2x - 3 = 0 \):
\[ 2x = 3 \implies x = \frac{3}{2} \]
From \( x - 1 = 0 \):
\[ x = 1 \]

**Final Answer for (i):**
The solutions are \( x = 1 \) and \( x = \frac{3}{2} \).

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### (ii) Solve: \( (x^2 + 3x)^2 - (x^2 + 3x) - 6 = 0 \)

**Step 1: Substitute \( y = x^2 + 3x \).**
Let \( y = x^2 + 3x \). The equation becomes:
\[ y^2 - y - 6 = 0 \]

**Step 2: Factor the quadratic equation.**
We need two numbers that multiply to -6 and add to -1. The numbers are -3 and 2.

**Step 3: Rewrite the equation using these numbers.**
\[ y^2 - 3y + 2y - 6 = 0 \]

**Step 4: Group the terms.**
Group the first two and the last two terms:
\[ (y^2 - 3y) + (2y - 6) = 0 \]

**Step 5: Factor by grouping.**
Factor out the common terms:
\[ y(y - 3) + 2(y - 3) = 0 \]
Now, factor out \( (y - 3) \):
\[ (y - 3)(y + 2) = 0 \]

**Step 6: Set each factor to zero.**
1. \( y - 3 = 0 \)
2. \( y + 2 = 0 \)

**Step 7: Solve for \( y \).**
From \( y - 3 = 0 \):
\[ y = 3 \]
From \( y + 2 = 0 \):
\[ y = -2 \]

**Step 8: Substitute back to find \( x \).**
Now we substitute back \( y = x^2 + 3x \):
1. For \( y = 3 \):
\[ x^2 + 3x - 3 = 0 \]
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = 3, c = -3 \):
\[ D = b^2 - 4ac = 3^2 - 4 \cdot 1 \cdot (-3) = 9 + 12 = 21 \]
Thus,
\[ x = \frac{-3 \pm \sqrt{21}}{2} \]

2. For \( y = -2 \):
\[ x^2 + 3x + 2 = 0 \]
This factors to:
\[ (x + 1)(x + 2) = 0 \]
So,
\[ x = -1 \quad \text{or} \quad x = -2 \]

**Final Answer for (ii):**
The solutions are \( x = -1, x = -2, x = \frac{-3 + \sqrt{21}}{2}, x = \frac{-3 - \sqrt{21}}{2} \).

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