5x=6x+38:6+x

To solve the equation \(5x = \frac{6x + 38}{6 + x}\), we will first eliminate the fraction by multiplying both sides by \(6 + x\):

\[ 5x(6 + x) = 6x + 38 \]

Now, expand both sides:

\[ 30x + 5x^2 = 6x + 38 \]

Next, rearrange the equation to set it to 0:

\[ 5x^2 + 30x - 6x - 38 = 0 \]

Combine like terms:

\[ 5x^2 + 24x - 38 = 0 \]

Now, we can use the quadratic formula to solve for \(x\). The quadratic formula is given by:

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

In our case, \(a = 5\), \(b = 24\), and \(c = -38\). First, we need to calculate the discriminant \(b^2 - 4ac\):

\[ b^2 - 4ac = 24^2 - 4(5)(-38) \] \[ = 576 + 760 \] \[ = 1336 \]

Now apply the quadratic formula:

\[ x = \frac{-24 \pm \sqrt{1336}}{2 \cdot 5} \]

Calculate \(\sqrt{1336}\):

\(\sqrt{1336} \approx 36.5\)

Now substitute back into the formula:

\[ x = \frac{-24 \pm 36.5}{10} \]

Calculating the two possible values of \(x\):

1. \(x = \frac{-24 + 36.5}{10} = \frac{12.5}{10} = 1.25\)

2. \(x = \frac{-24 - 36.5}{10} = \frac{-60.5}{10} = -6.05\)

Thus, the solutions to the equation are:

\[ x \approx 1.25 \quad \text{and} \quad x \approx -6.05 \]