Solve the equation x^(2)-3x+9=-10 x to the nearest tenth. Answer: x=

Set Equation to Zero: First, we need to set the equation to zero by adding 10 to both sides of the equation. x^2 - 3x + 9 = -10 x^2 - 3x + 9 + 10 = 0 x^2 - 3x + 19 = 0Use Quadratic Formula: Next, we will use the quadratic formula to solve for x. The quadratic formula is x = (-b ± √(b^2 - 4ac)) / (2a), where a, b, and c are the coefficients from the quadratic equation ax^2 + bx + c = 0. In our equation, a = 1, b = -3, and c = 19.Calculate Discriminant: Now we will calculate the discriminant, which is the part under the square root in the quadratic formula: b^2 - 4ac. Discriminant = (-3)^2 - 4(1)(19) Discriminant = 9 - 76 Discriminant = -67Identify Complex Solutions: Since the discriminant is negative, there are no real solutions to the equation. The solutions are complex numbers. Therefore, we cannot round to the nearest tenth as the question prompt asks for real number solutions.

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Solve the equation
x^(2)-3x+9=-10 x to the nearest tenth.
Answer:
x=

Solve the equation $x^{2}-3 x+9=-10 x$ to the nearest tenth. $\newline$ Answer: $x=$

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Q. Solve the equation

x^{2}-3 x+9=-10 x

to the nearest tenth.

\newline

Answer:

x=

Set Equation to Zero: First, we need to set the equation to zero by adding $10$ to both sides of the equation. $\newline$ $x^2 - 3x + 9 = -10$ $\newline$ $x^2 - 3x + 9 + 10 = 0$ $\newline$ $x^2 - 3x + 19 = 0$
Use Quadratic Formula: Next, we will use the quadratic formula to solve for $x$ . The quadratic formula is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ , where $a$ , $b$ , and $c$ are the coefficients from the quadratic equation $ax^2 + bx + c = 0$ . In our equation, $a = 1$ , $b = -3$ , and $c = 19$ .
Calculate Discriminant: Now we will calculate the discriminant, which is the part under the square root in the quadratic formula: $b^2 - 4ac$ .
Discriminant = $(-3)^2 - 4(1)(19)$
Discriminant = $9 - 76$
Discriminant = $-67$
Identify Complex Solutions: Since the discriminant is negative, there are no real solutions to the equation. The solutions are complex numbers. $\newline$ Therefore, we cannot round to the nearest tenth as the question prompt asks for real number solutions.