Solve for x. Enter the solutions from least to greatest. 2x^(2)-5=13 lesser x= greater x=

Isolate x^2 term: The question prompt is: ＂Solve for x where 2x^2 - 5 = 13.＂ First, we need to isolate the x^2 term by moving the constant term to the other side of the equation. 2x^2 - 5 + 5 = 13 + 5 2x^2 = 18Divide by 2: Next, we divide both sides of the equation by 2 to solve for x^2. 2x^2 / 2 = 18 / 2 x^2 = 9Take 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 = ±√9 x = ±3Two solutions for x: We have found two solutions for x. The lesser value is -3 and the greater value is 3. lesser x = -3 greater x = 3

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

2x^(2)-5=13
lesser
x=
greater
x=

Solve for $x$ . $\newline$ Enter the solutions from least to greatest. $\newline$ $2 x^{2}-5=13$ $\newline$ 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

2 x^{2}-5=13

\newline

lesser

x=

\newline

greater

x=

Isolate $x^2$ term: The question prompt is: ＂Solve for $x$ where $2x^2 - 5 = 13$ .＂ $\newline$ First, we need to isolate the $x^2$ term by moving the constant term to the other side of the equation. $\newline$ $2x^2 - 5 + 5 = 13 + 5$ $\newline$ $2x^2 = 18$
Divide by $2$ : Next, we divide both sides of the equation by $2$ to solve for $x^2$ . $\newline$ $\frac{2x^2}{2} = \frac{18}{2}$ $\newline$ $x^2 = 9$
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. $\newline$ $\sqrt{x^2} = \pm\sqrt{9}$ $\newline$ x = $\pm3$
Two solutions for x: We have found two solutions for x. The lesser value is $-3$ and the greater value is $3$ . $\newline$ lesser $x = -3$ $\newline$ greater $x = 3$