For which quadratic equation is $$ 2-\sqrt{3} $$ a solution? a) $$ x^{2}-4 x=3 $$ b) $$ x^{2}-2 x=3 $$ c) $$ x^{2}-2 x=2 $$ d) $$ x^{2}-4 x=-1 $$

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To determine which quadratic equation has $$ 2 - \sqrt{3} $$ as a solution, we will substitute $$ x = 2 - \sqrt{3} $$ into each equation and check if the equation holds true.

Let's evaluate each option step by step:

1. **Option a:** $$ x^{2} - 4x = 3 $$
   - Substitute $$ x = 2 - \sqrt{3} $$:
   $$
   (2 - \sqrt{3})^{2} - 4(2 - \sqrt{3}) = 3
   $$

2. **Option b:** $$ x^{2} - 2x = 3 $$
   - Substitute $$ x = 2 - \sqrt{3} $$:
   $$
   (2 - \sqrt{3})^{2} - 2(2 - \sqrt{3}) = 3
   $$

3. **Option c:** $$ x^{2} - 2x = 2 $$
   - Substitute $$ x = 2 - \sqrt{3} $$:
   $$
   (2 - \sqrt{3})^{2} - 2(2 - \sqrt{3}) = 2
   $$

4. **Option d:** $$ x^{2} - 4x = -1 $$
   - Substitute $$ x = 2 - \sqrt{3} $$:
   $$
   (2 - \sqrt{3})^{2} - 4(2 - \sqrt{3}) = -1
   $$

Now, let's calculate each substitution to find out which equation holds true.

First, we will calculate $$ (2 - \sqrt{3})^{2} $$ and $$ 4(2 - \sqrt{3}) $$ for all options.

Let's compute $$ (2 - \sqrt{3})^{2} $$:
Calculate the value by following steps:
- step0: Calculate:
  $$\left(2-\sqrt{3}\right)^{2}$$
- step1: Expand the expression:
  $$2^{2}-2\times 2\sqrt{3}+\left(\sqrt{3}\right)^{2}$$
- step2: Calculate:
  $$4-4\sqrt{3}+3$$
- step3: Add the numbers:
  $$7-4\sqrt{3}$$
The calculation of $$ (2 - \sqrt{3})^{2} $$ gives us $$ 7 - 4\sqrt{3} $$.

Now, we will compute $$ 4(2 - \sqrt{3}) $$:
Calculate the value by following steps:
- step0: Calculate:
  $$4\left(2-\sqrt{3}\right)$$
- step1: Apply the distributive property:
  $$4\times 2-4\sqrt{3}$$
- step2: Multiply the numbers:
  $$8-4\sqrt{3}$$
The calculation of $$ 4(2 - \sqrt{3}) $$ gives us $$ 8 - 4\sqrt{3} $$.

Now, we can evaluate each option:

1. **Option a:** $$ (2 - \sqrt{3})^{2} - 4(2 - \sqrt{3}) = 3 $$
   $$
   (7 - 4\sqrt{3}) - (8 - 4\sqrt{3}) = 3 \implies 7 - 8 = 3 \implies -1 \neq 3
   $$

2. **Option b:** $$ (2 - \sqrt{3})^{2} - 2(2 - \sqrt{3}) = 3 $$
   - First, calculate $$ 2(2 - \sqrt{3}) $$:
   $$
   2(2 - \sqrt{3}) = 4 - 2\sqrt{3}
   $$
   - Now substitute:
   $$
   (7 - 4\sqrt{3}) - (4 - 2\sqrt{3}) = 3 \implies 7 - 4\sqrt{3} - 4 + 2\sqrt{3} = 3 \implies 3 - 2\sqrt{3} \neq 3
   $$

3. **Option c:** $$ (2 - \sqrt{3})^{2} - 2(2 - \sqrt{3}) = 2 $$
   - We already have $$ 2(2 - \sqrt{3}) = 4 - 2\sqrt{3} $$:
   $$
   (7 - 4\sqrt{3}) - (4 - 2\sqrt{3}) = 2 \implies 7 - 4\sqrt{3} - 4 + 2\sqrt{3} = 2 \implies 3 - 2\sqrt{3} \neq 2
   $$

4. **Option d:** $$ (2 - \sqrt{3})^{2} - 4(2 - \sqrt{3}) = -1 $$
   - We already have $$ 4(2 - \sqrt{3}) = 8 - 4\sqrt{3} $$:
   $$
   (7 - 4\sqrt{3}) - (8 - 4\sqrt{3}) = -1 \implies 7 - 8 = -1 \implies -1 = -1
   $$

The only equation that holds true is **Option d:** $$ x^{2} - 4x = -1 $$.

Thus, the quadratic equation for which $$ 2 - \sqrt{3} $$ is a solution is:
**d) $$ x^{2} - 4x = -1 $$**.
The quadratic equation where $$ 2 - \sqrt{3} $$ is a solution is: d) $$ x^{2} - 4x = -1 $$.

For which quadratic equation is a solution?
a)
b)
c)
d)

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