Find the value of 'x' and 'y' from the following matrix equation :
`[(2x+1,2y),(0,y^(2)-5y)]=[(x+3,y^(2)+2),(0,-6)]`

Answer

To solve the matrix equation

\[ \begin{pmatrix} 2x + 1 & 2y \\ 0 & y^2 - 5y \end{pmatrix} = \begin{pmatrix} x + 3 & y^2 + 2 \\ 0 & -6 \end{pmatrix} \]

we will equate the corresponding elements of the matrices.

### Step 1: Equate the first elements
From the first elements of the matrices, we have:

\[ 2x + 1 = x + 3 \]

### Step 2: Solve for \(x\)
Rearranging the equation:

\[ 2x - x = 3 - 1 \]
\[ x = 2 \]

### Step 3: Equate the second elements
Now, equate the second elements of the matrices:

\[ 2y = y^2 + 2 \]

### Step 4: Rearrange the equation
Rearranging gives:

\[ y^2 - 2y + 2 = 0 \]

### Step 5: Solve the quadratic equation
Now we will use the quadratic formula to solve for \(y\):

The quadratic formula is given by:

\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Here, \(a = 1\), \(b = -2\), and \(c = 2\).

Calculating the discriminant:

\[ b^2 - 4ac = (-2)^2 - 4 \cdot 1 \cdot 2 = 4 - 8 = -4 \]

Since the discriminant is negative, there are no real solutions for \(y\).

### Step 6: Conclusion
Thus, we have found:

\[ x = 2 \]

But there are no real values for \(y\) that satisfy the equation.

### Summary
The values are:
- \(x = 2\)
- No real values for \(y\)