Find the zeros of the function. Enter the solutions from least to greatest. f(x)=-3x^(2)+75 lesser x= greater x=

Move constant term: Now, we will move the constant term to the other side of the equation by adding 3x^2 to both sides. 0 + 3x^2 = -3x^2 + 75 + 3x^2 3x^2 = 75Divide to isolate x^2: Next, we will divide both sides of the equation by 3 to isolate x^2. (3x^2) / 3 = 75 / 3 x^2 = 25Take square root: To find the values of x, we take the square root of both sides. Remember that taking the square root of a number yields both a positive and a negative solution. x = ±√25 x = ±5Find solutions for x: We have found two solutions for x. The lesser value is -5 and the greater value is 5. lesser x = -5 greater x = 5

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Find the zeros of the function.
Enter the solutions from least to greatest.

f(x)=-3x^(2)+75
lesser
x=
greater
x=

Find the zeros of the function. $\newline$ Enter the solutions from least to greatest. $\newline$ $f(x)=-3 x^{2}+75$ $\newline$ lesser $x=$ $\newline$ greater $x=$

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Math Problems

Algebra 1

Evaluate recursive formulas for sequences

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Q. Find the zeros of the function.

\newline

Enter the solutions from least to greatest.

\newline

f(x)=-3 x^{2}+75

\newline

lesser

x=

\newline

greater

x=

Move constant term: Now, we will move the constant term to the other side of the equation by adding $3x^2$ to both sides. $\newline$ $0 + 3x^2 = -3x^2 + 75 + 3x^2$ $\newline$ $3x^2 = 75$
Divide to isolate $x^2$ : Next, we will divide both sides of the equation by $3$ to isolate $x^2$ . $\newline$ $\frac{3x^2}{3} = \frac{75}{3}$ $\newline$ $x^2 = 25$
Take square root: To find the values of $x$ , we take the square root of both sides. Remember that taking the square root of a number yields both a positive and a negative solution. $\newline$ $x = \pm\sqrt{25}$ $\newline$ $x = \pm5$
Find solutions for $x$ : We have found two solutions for $x$ . The lesser value is $-5$ and the greater value is $5$ . $\newline$ lesser $x$ = $-5$ $\newline$ greater $x$ = $5$