Solve for x. Enter the solutions from least to greatest. 5x^(2)+3=83 lesser x= greater x=

Isolate x term: First, we need to isolate the term with the variable x on one side of the equation. We start by subtracting 3 from both sides of the equation to get rid of the constant term on the left side. 5x^2 + 3 = 83 5x^2 = 83 - 3 5x^2 = 80Divide by 5: Next, we divide both sides of the equation by 5 to solve for x^2. 5x^2 / 5 = 80 / 5 x^2 = 16Take square root: Now, we take the square root of both sides to solve for x. Remember that taking the square root of a number yields two solutions: one positive and one negative. √(x^2) = ±√(16) x = ±4List solutions: We have two solutions for x: x = 4 and x = -4. We need to list them from least to greatest. lesser x = -4 greater x = 4

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Solve for
x.
Enter the solutions from least to greatest.

5x^(2)+3=83
lesser
x=
greater
x=

Solve for $x$ . $\newline$ Enter the solutions from least to greatest. $\newline$ $5 x^{2}+3=83$ $\newline$ lesser $x=$ $\newline$ greater $x=$

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Math Problems

Algebra 1

Evaluate recursive formulas for sequences

Full solution

Q. Solve for

x

\newline

Enter the solutions from least to greatest.

\newline

5 x^{2}+3=83

\newline

lesser

x=

\newline

greater

x=

Isolate x term: First, we need to isolate the term with the variable $x$ on one side of the equation. We start by subtracting $3$ from both sides of the equation to get rid of the constant term on the left side. $\newline$ $5x^2 + 3 = 83$ $\newline$ $5x^2 = 83 - 3$ $\newline$ $5x^2 = 80$
Divide by $5$ : Next, we divide both sides of the equation by $5$ to solve for $x^2$ . $\newline$ $\frac{5x^2}{5} = \frac{80}{5}$ $\newline$ $x^2 = 16$
Take square root: Now, we take the square root of both sides to solve for x. Remember that taking the square root of a number yields two solutions: one positive and one negative. $\newline$ $\sqrt{x^2} = \pm\sqrt{16}$ $\newline$ $x = \pm4$
List solutions: We have two solutions for x: $x = 4$ and $x = -4$ . We need to list them from least to greatest. $\newline$ lesser $x = -4$ $\newline$ greater $x = 4$