x2-12x+.35=0

To solve the quadratic equation \(x^2 - 12x + 0.35 = 0\), we can use the quadratic formula:

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

In this case, the coefficients are:

• \(a = 1\)
• \(b = -12\)
• \(c = 0.35\)

Now, we can plug these values into the formula:

1. Calculate the discriminant \(b^2 - 4ac\):

\[ b^2 = (-12)^2 = 144 \] \[ 4ac = 4 \cdot 1 \cdot 0.35 = 1.4 \] \[ b^2 - 4ac = 144 - 1.4 = 142.6 \]

2. Now substitute into the quadratic formula:

\[ x = \frac{-(-12) \pm \sqrt{142.6}}{2 \cdot 1} \] \[ x = \frac{12 \pm \sqrt{142.6}}{2} \]

3. Calculate \(\sqrt{142.6}\):

\[ \sqrt{142.6} \approx 11.92 \text{ (approximately)} \]

4. Substitute back into the equation:

\[ x = \frac{12 \pm 11.92}{2} \]

This gives us two solutions:

1. \(x_1 = \frac{12 + 11.92}{2} = \frac{23.92}{2} \approx 11.96\)
2. \(x_2 = \frac{12 - 11.92}{2} = \frac{0.08}{2} \approx 0.04\)

Therefore, the solutions to the equation \(x^2 - 12x + 0.35 = 0\) are approximately:

\[ x \approx 11.96 \quad \text{and} \quad x \approx 0.04 \]