For what values of a and b, the equation `x^(2)+(2a-3)x=3b+4` should have both the roots zero?

Answer

To solve the problem, we need to determine the values of \( a \) and \( b \) such that the quadratic equation

\[ x^2 + (2a - 3)x = 3b + 4 \]

has both roots equal to zero.

### Step-by-Step Solution:

1. **Rearrange the Equation**:
We start with the given equation:
\[ x^2 + (2a - 3)x - (3b + 4) = 0 \]

2. **Identify Conditions for Roots**:
For both roots of the quadratic equation to be zero, the coefficients of \( x \) and the constant term must both be zero. This gives us two equations:
- The coefficient of \( x \) (which is \( 2a - 3 \)) must be zero:
\[ 2a - 3 = 0 \]
- The constant term (which is \( -(3b + 4) \)) must also be zero:
\[ -(3b + 4) = 0 \]

3. **Solve for \( a \)**:
From the first equation:
\[ 2a - 3 = 0 \implies 2a = 3 \implies a = \frac{3}{2} \]

4. **Solve for \( b \)**:
From the second equation:
\[ -(3b + 4) = 0 \implies 3b + 4 = 0 \implies 3b = -4 \implies b = -\frac{4}{3} \]

5. **Final Values**:
Therefore, the values of \( a \) and \( b \) are:
\[ a = \frac{3}{2}, \quad b = -\frac{4}{3} \]

### Conclusion:
The values of \( a \) and \( b \) such that the equation \( x^2 + (2a - 3)x = 3b + 4 \) has both roots equal to zero are:
\[ \boxed{a = \frac{3}{2}, \quad b = -\frac{4}{3}} \]