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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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Solve exponential equations using logarithms

Full solution

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.

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