(2x+3)(3x+1)=11x-30

To solve the equation \( (2x+3)(3x+1)=11x-30 \), let's first expand the left-hand side.

\[ (2x+3)(3x+1) = 2x \cdot 3x + 2x \cdot 1 + 3 \cdot 3x + 3 \cdot 1 \] \[ = 6x^2 + 2x + 9x + 3 \] Combine like terms: \[ = 6x^2 + 11x + 3 \]

Now, we rewrite the equation with the expanded left-hand side:

\[ 6x^2 + 11x + 3 = 11x - 30 \]

Next, we can subtract \( 11x \) from both sides:

\[ 6x^2 + 3 = -30 \]

Now, let's add \( 30 \) to both sides:

\[ 6x^2 + 3 + 30 = 0 \] \[ 6x^2 + 33 = 0 \]

Now we can simplify this by subtracting \( 33 \) from both sides:

\[ 6x^2 = -33 \]

Dividing by \( 6 \):

\[ x^2 = -\frac{33}{6} \]

This simplifies to:

\[ x^2 = -\frac{11}{2} \]

Since \( x^2 \) cannot be negative in the set of real numbers, this equation has no real solutions. Therefore, the solutions are complex:

\[ x = \pm i \sqrt{\frac{11}{2}} \]

In summary, the solutions to the equation are:

\[ x = \frac{\sqrt{22}}{2} i \quad \text{and} \quad x = -\frac{\sqrt{22}}{2} i \]