To solve the quadratic equation \(-x^2 + x = 10\), we first rearrange it to the standard form \(ax^2 + bx + c = 0\):
\[ -x^2 + x - 10 = 0 \]
Now, let's multiply through by \(-1\) to make the leading coefficient positive:
\[ x^2 - x + 10 = 0 \]
Next, we can use the quadratic formula to find the solutions. The quadratic formula is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In this equation, \(a = 1\), \(b = -1\), and \(c = 10\). We plug these values into the formula:
\[ x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot 10}}{2 \cdot 1} \] \[ x = \frac{1 \pm \sqrt{1 - 40}}{2} \] \[ x = \frac{1 \pm \sqrt{-39}}{2} \]
Since the discriminant (\(b^2 - 4ac = -39\)) is negative, this indicates that there are no real solutions, but rather two complex solutions.
To express the complex solutions, we can simplify \(\sqrt{-39}\):
\[ \sqrt{-39} = i\sqrt{39} \]
Thus, the solutions can be written as:
\[ x = \frac{1 \pm i\sqrt{39}}{2} \]
Since there are no real solutions and you asked for a solution rounded to the nearest tenth, the answer is that the equation has no real solutions. The solutions are complex and cannot be rounded numerically in the context of real numbers.