Solve the equation -14 x+15=-x^(2) to the nearest tenth. Answer: x=

Rewrite equation in standard form: Write the given equation in standard quadratic form. To solve the quadratic equation, we need to rearrange it into the standard form ax^2 + bx + c = 0. The given equation is -14x + 15 = -x^2. We can rewrite it as x^2 - 14x + 15 = 0 by adding x^2 to both sides and rearranging the terms.Factor the quadratic equation: Factor the quadratic equation. We need to factor the quadratic equation x^2 - 14x + 15 = 0. The factors of 15 that add up to -14 are -1 and -15. So, we can write the equation as (x - 1)(x - 15) = 0.Solve using zero product property: Solve for x using the zero product property. If (x - 1)(x - 15) = 0, then either x - 1 = 0 or x - 15 = 0. Solving for x gives us x = 1 or x = 15.Round solutions: Round the solutions to the nearest tenth. The solutions x = 1 and x = 15 are already to the nearest tenth, so no further rounding is necessary.

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Solve the equation
-14 x+15=-x^(2) to the nearest tenth.
Answer:
x=

Solve the equation $-14 x+15=-x^{2}$ to the nearest tenth. $\newline$ Answer: $x=$

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Q. Solve the equation

-14 x+15=-x^{2}

to the nearest tenth.

\newline

Answer:

x=

Rewrite equation in standard form: Write the given equation in standard quadratic form. $\newline$ To solve the quadratic equation, we need to rearrange it into the standard form $ax^2 + bx + c = 0$ . $\newline$ The given equation is $-14x + 15 = -x^2$ . $\newline$ We can rewrite it as $x^2 - 14x + 15 = 0$ by adding $x^2$ to both sides and rearranging the terms.
Factor the quadratic equation: Factor the quadratic equation. $\newline$ We need to factor the quadratic equation $x^2 - 14x + 15 = 0$ . $\newline$ The factors of $15$ that add up to $-14$ are $-1$ and $-15$ . $\newline$ So, we can write the equation as $(x - 1)(x - 15) = 0$ .
Solve using zero product property: Solve for $x$ using the zero product property. $\newline$ If $(x - 1)(x - 15) = 0$ , then either $x - 1 = 0$ or $x - 15 = 0$ . $\newline$ Solving for $x$ gives us $x = 1$ or $x = 15$ .
Round solutions: Round the solutions to the nearest tenth. The solutions $x = 1$ and $x = 15$ are already to the nearest tenth, so no further rounding is necessary.