To solve the equation \( x^2 - 11x + 41 = 19 \), we first need to rearrange it into standard form. We'll subtract 19 from both sides:
\[ x^2 - 11x + 41 - 19 = 0 \]
This simplifies to:
\[ x^2 - 11x + 22 = 0 \]
Next, we can use the quadratic formula to solve for \( x \). The quadratic formula is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = -11 \), and \( c = 22 \).
Now, we can substitute these values into the formula:
- Calculate \( b^2 - 4ac \):
\[ (-11)^2 - 4(1)(22) = 121 - 88 = 33 \]
- Substitute into the quadratic formula:
\[ x = \frac{-(-11) \pm \sqrt{33}}{2(1)} = \frac{11 \pm \sqrt{33}}{2} \]
Next, we find the approximate value of \( \sqrt{33} \).
\[ \sqrt{33} \approx 5.744 \]
Now, substituting this value back into the equation:
\[ x = \frac{11 \pm 5.744}{2} \]
Calculating the two possible values for \( x \):
- First solution:
\[ x_1 = \frac{11 + 5.744}{2} = \frac{16.744}{2} \approx 8.372 \]
- Second solution:
\[ x_2 = \frac{11 - 5.744}{2} = \frac{5.256}{2} \approx 2.628 \]
Thus, the solutions for \( x \) rounded to the nearest tenth are:
\[ x \approx 8.4 \quad \text{and} \quad x \approx 2.6 \]