Solve the Quadratic Equation: -x² + 6x - 10 = 0

This is a quadratic equation:

$-x^2+6x-10=0$

due to the fact that there is a quadratic term present(meaning raised to the second power),

The first step in solving a quadratic equation is always arranging it in a form where all terms on one side are ordered from highest to lowest power (in descending order from left to right) and 0 on the other side,

Then we can choose whether to solve it using the quadratic formula or by factoring/completing the square.

The equation in the problem is already arranged, so let's proceed with the solving technique:

For ease of solving and minimizing errors, it is always recommended to ensure that the coefficient of the quadratic term in the equation is positive,

We'll achieve this by multiplying (both sides of) the equation by:$-1$:

$-x^2+6x-10=0 \hspace{8pt}\text{/}\cdot(-1)\\ x^2-6x+10=0$$$

Let's continue solving the equation

Solve it using the quadratic formula,

Remember:

The rule states that the roots of an equation of the form:

$ax^2+bx+c=0$

are:

$x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

(meaning its solutions, the two possible values of the unknown for which we get a true statement when substituted in the equation)

$$This formula is called: "The Quadratic Formula"

Let's return to the problem:

$x^2-6x+10=0$

and solve it:

First, let's identify the coefficients of the terms:

$\begin{cases}a=1 \\ b=-6 \\ c=10\end{cases}$

where we noted that the coefficient of the quadratic term is 1,

We obtain the equation's solutions (roots) by substituting these coefficients that we mentioned earlier in the quadratic formula:

$x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-(-6)\pm\sqrt{(-6)^2-4\cdot1\cdot10}}{2\cdot1}$

Let's continue and calculate the expression under the root and simplify the expression:

$x_{1,2}=\frac{6\pm\sqrt{-4}}{2}$

The expression under the root is negative, and since we cannot extract a real root from a negative number, this equation has no real solutions,

Meaning - there is no real value of $x$that when substituted in the equation will give a true statement.

Therefore, the correct answer is answer D.