Solve by Factoring 2x = square root of 20+2x

$2 x = \sqrt{20 + 2 x}$

Move $\sqrt{20 + 2 x}$ to the left side of the equation by subtracting it from both sides.

$2 x - \sqrt{20 + 2 x} = 0$

Factor $2$ out of $20 + 2 x$ .

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Factor $2$ out of $20$ .

$2 x - \sqrt{2 \cdot 10 + 2 x} = 0$

Factor $2$ out of $2 \cdot 10 + 2 x$ .

$2 x - \sqrt{2 (10 + x)} = 0$

Subtract $2 x$ from both sides of the equation.

$- \sqrt{2 (10 + x)} = - 2 x$

Multiply each term in $- \sqrt{2 (10 + x)} = - 2 x$ by $- 1$

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Multiply each term in $- \sqrt{2 (10 + x)} = - 2 x$ by $- 1$ .

$(- \sqrt{2 (10 + x)}) \cdot - 1 = (- 2 x) \cdot - 1$

Multiply $- \sqrt{2 (10 + x)} \cdot - 1$ .

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Multiply $- 1$ by $- 1$ .

$1 \sqrt{2 (10 + x)} = (- 2 x) \cdot - 1$

Multiply $\sqrt{2 (10 + x)}$ by $1$ .

$\sqrt{2 (10 + x)} = (- 2 x) \cdot - 1$

Multiply $- 1$ by $- 2$ .

$\sqrt{2 (10 + x)} = 2 x$

To remove the radical on the left side of the equation, square both sides of the equation.

${\sqrt{2 (10 + x)}}^{2} = {(2 x)}^{2}$

Simplify each side of the equation.

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Multiply the exponents in ${({(2 (10 + x))}^{\frac{1}{2}})}^{2}$ .

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Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

${(2 (10 + x))}^{\frac{1}{2} \cdot 2} = {(2 x)}^{2}$

Cancel the common factor of $2$ .

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Cancel the common factor.

${(2 (10 + x))}^{\frac{1}{2} \cdot 2} = {(2 x)}^{2}$

Rewrite the expression.

${(2 (10 + x))}^{1} = {(2 x)}^{2}$

Apply the distributive property.

${(2 \cdot 10 + 2 x)}^{1} = {(2 x)}^{2}$

Multiply $2$ by $10$ .

${(20 + 2 x)}^{1} = {(2 x)}^{2}$

Simplify.

$20 + 2 x = {(2 x)}^{2}$

Apply the product rule to $2 x$ .

$20 + 2 x = 2^{2} x^{2}$

Raise $2$ to the power of $2$ .

$20 + 2 x = 4 x^{2}$

Solve for $x$ .

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Subtract $4 x^{2}$ from both sides of the equation.

$20 + 2 x - 4 x^{2} = 0$

Factor the left side of the equation.

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Factor $- 2$ out of $20 + 2 x - 4 x^{2}$ .

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Reorder the expression.

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Move $20$ .

$2 x - 4 x^{2} + 20 = 0$

Reorder $2 x$ and $- 4 x^{2}$ .

$- 4 x^{2} + 2 x + 20 = 0$

Factor $- 2$ out of $- 4 x^{2}$ .

$- 2 (2 x^{2}) + 2 x + 20 = 0$

Factor $- 2$ out of $2 x$ .

$- 2 (2 x^{2}) - 2 (- x) + 20 = 0$

Factor $- 2$ out of $20$ .

$- 2 (2 x^{2}) - 2 (- x) - 2 \cdot - 10 = 0$

Factor $- 2$ out of $- 2 (2 x^{2}) - 2 (- x)$ .

$- 2 (2 x^{2} - x) - 2 \cdot - 10 = 0$

Factor $- 2$ out of $- 2 (2 x^{2} - x) - 2 (- 10)$ .

$- 2 (2 x^{2} - x - 10) = 0$

Factor.

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Factor by grouping.

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For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot - 10 = - 20$ and whose sum is $b = - 1$ .

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Factor $- 1$ out of $- x$ .

$- 2 (2 x^{2} - (x) - 10) = 0$

Rewrite $- 1$ as $4$ plus $- 5$

$- 2 (2 x^{2} + (4 - 5) x - 10) = 0$

Apply the distributive property.

$- 2 (2 x^{2} + 4 x - 5 x - 10) = 0$

Factor out the greatest common factor from each group.

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Group the first two terms and the last two terms.

$- 2 ((2 x^{2} + 4 x) - 5 x - 10) = 0$

Factor out the greatest common factor (GCF) from each group.

$- 2 (2 x (x + 2) - 5 (x + 2)) = 0$

Factor the polynomial by factoring out the greatest common factor, $x + 2$ .

$- 2 ((x + 2) (2 x - 5)) = 0$

Remove unnecessary parentheses.

$- 2 (x + 2) (2 x - 5) = 0$

Divide each term by $- 2$ and simplify.

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Divide each term in $- 2 (x + 2) (2 x - 5) = 0$ by $- 2$ .

$\frac{- 2 (x + 2) (2 x - 5)}{- 2} = \frac{0}{- 2}$

Simplify $\frac{- 2 (x + 2) (2 x - 5)}{- 2}$ .

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Cancel the common factor of $- 2$ .

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Cancel the common factor.

$\frac{- 2 (x + 2) (2 x - 5)}{- 2} = \frac{0}{- 2}$

Divide $(x + 2) (2 x - 5)$ by $1$ .

$(x + 2) (2 x - 5) = \frac{0}{- 2}$

Expand $(x + 2) (2 x - 5)$ using the FOIL Method.

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Apply the distributive property.

$x (2 x - 5) + 2 (2 x - 5) = \frac{0}{- 2}$

Apply the distributive property.

$x (2 x) + x \cdot - 5 + 2 (2 x - 5) = \frac{0}{- 2}$

Apply the distributive property.

$x (2 x) + x \cdot - 5 + 2 (2 x) + 2 \cdot - 5 = \frac{0}{- 2}$

Simplify and combine like terms.

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Simplify each term.

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Rewrite using the commutative property of multiplication.

$2 x \cdot x + x \cdot - 5 + 2 (2 x) + 2 \cdot - 5 = \frac{0}{- 2}$

Multiply $x$ by $x$ by adding the exponents.

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Move $x$ .

$2 (x \cdot x) + x \cdot - 5 + 2 (2 x) + 2 \cdot - 5 = \frac{0}{- 2}$

Multiply $x$ by $x$ .

$2 x^{2} + x \cdot - 5 + 2 (2 x) + 2 \cdot - 5 = \frac{0}{- 2}$

Move $- 5$ to the left of $x$ .

$2 x^{2} - 5 \cdot x + 2 (2 x) + 2 \cdot - 5 = \frac{0}{- 2}$

Multiply $2$ by $2$ .

$2 x^{2} - 5 x + 4 x + 2 \cdot - 5 = \frac{0}{- 2}$

Multiply $2$ by $- 5$ .

$2 x^{2} - 5 x + 4 x - 10 = \frac{0}{- 2}$

Add $- 5 x$ and $4 x$ .

$2 x^{2} - x - 10 = \frac{0}{- 2}$

Divide $0$ by $- 2$ .

$2 x^{2} - x - 10 = 0$

Factor by grouping.

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For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot - 10 = - 20$ and whose sum is $b = - 1$ .

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Factor $- 1$ out of $- x$ .

$2 x^{2} - (x) - 10 = 0$

Rewrite $- 1$ as $4$ plus $- 5$

$2 x^{2} + (4 - 5) x - 10 = 0$

Apply the distributive property.

$2 x^{2} + 4 x - 5 x - 10 = 0$

Factor out the greatest common factor from each group.

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Group the first two terms and the last two terms.

$(2 x^{2} + 4 x) - 5 x - 10 = 0$

Factor out the greatest common factor (GCF) from each group.

$2 x (x + 2) - 5 (x + 2) = 0$

Factor the polynomial by factoring out the greatest common factor, $x + 2$ .

$(x + 2) (2 x - 5) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x + 2 = 0$

$2 x - 5 = 0$

Set the first factor equal to $0$ and solve.

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Set the first factor equal to $0$ .

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

Set the next factor equal to $0$ and solve.

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Set the next factor equal to $0$ .

$2 x - 5 = 0$

Add $5$ to both sides of the equation.

$2 x = 5$

Divide each term by $2$ and simplify.

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Divide each term in $2 x = 5$ by $2$ .

$\frac{2 x}{2} = \frac{5}{2}$

Cancel the common factor of $2$ .

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Cancel the common factor.

$\frac{2 x}{2} = \frac{5}{2}$

Divide $x$ by $1$ .

$x = \frac{5}{2}$

The final solution is all the values that make $(x + 2) (2 x - 5) = 0$ true.

$x = - 2, \frac{5}{2}$

Exclude the solutions that do not make $2 x - \sqrt{2 (10 + x)} = 0$ true.

$x = \frac{5}{2}$

Solve by Factoring 2x = square root of 20+2x

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