Pre-Algebra Examples

$x^{2} + 50 = 25 + 2 x^{2}$

Step 1

Move all terms containing $x$ to the left side of the equation.

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Step 1.1

Subtract $2 x^{2}$ from both sides of the equation.

$x^{2} + 50 - 2 x^{2} = 25$

Step 1.2

Subtract $2 x^{2}$ from $x^{2}$ .

$- x^{2} + 50 = 25$

Step 2

Move all terms not containing $x$ to the right side of the equation.

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Step 2.1

Subtract $50$ from both sides of the equation.

$- x^{2} = 25 - 50$

Step 2.2

Subtract $50$ from $25$ .

$- x^{2} = - 25$

Step 3

Divide each term in $- x^{2} = - 25$ by $- 1$ and simplify.

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Step 3.1

Divide each term in $- x^{2} = - 25$ by $- 1$ .

$\frac{- x^{2}}{- 1} = \frac{- 25}{- 1}$

Step 3.2

Simplify the left side.

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Step 3.2.1

Dividing two negative values results in a positive value.

$\frac{x^{2}}{1} = \frac{- 25}{- 1}$

Step 3.2.2

Divide $x^{2}$ by $1$ .

$x^{2} = \frac{- 25}{- 1}$

Step 3.3

Simplify the right side.

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Step 3.3.1

Divide $- 25$ by $- 1$ .

$x^{2} = 25$

Step 4

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{25}$

Step 5

Simplify $\pm \sqrt{25}$ .

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Step 5.1

Rewrite $25$ as $5^{2}$ .

$x = \pm \sqrt{5^{2}}$

Step 5.2

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 5$

Step 6

The complete solution is the result of both the positive and negative portions of the solution.

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Step 6.1

First, use the positive value of the $\pm$ to find the first solution.

$x = 5$

Step 6.2

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 5$

Step 6.3

The complete solution is the result of both the positive and negative portions of the solution.

$x = 5, - 5$