Solve by Factoring 28x^2+7=28x

$28 x^{2} + 7 = 28 x$

Step 1

Subtract $28 x$ from both sides of the equation.

$28 x^{2} + 7 - 28 x = 0$

Step 2

Factor $7$ out of $28 x^{2} + 7 - 28 x$ .

Factor $7$ out of $28 x^{2}$ .

$7 (4 x^{2}) + 7 - 28 x = 0$

Factor $7$ out of $7$ .

$7 (4 x^{2}) + 7 (1) - 28 x = 0$

Factor $7$ out of $- 28 x$ .

$7 (4 x^{2}) + 7 (1) + 7 (- 4 x) = 0$

Factor $7$ out of $7 (4 x^{2}) + 7 (1)$ .

$7 (4 x^{2} + 1) + 7 (- 4 x) = 0$

Factor $7$ out of $7 (4 x^{2} + 1) + 7 (- 4 x)$ .

$7 (4 x^{2} + 1 - 4 x) = 0$

Step 3

Factor using the perfect square rule.

Rearrange terms.

$7 (4 x^{2} - 4 x + 1) = 0$

Rewrite $4 x^{2}$ as ${(2 x)}^{2}$ .

$7 ({(2 x)}^{2} - 4 x + 1) = 0$

Rewrite $1$ as $1^{2}$ .

$7 ({(2 x)}^{2} - 4 x + 1^{2}) = 0$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$4 x = 2 \cdot (2 x) \cdot 1$

Rewrite the polynomial.

$7 ({(2 x)}^{2} - 2 \cdot (2 x) \cdot 1 + 1^{2}) = 0$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = 2 x$ and $b = 1$ .

$7 {(2 x - 1)}^{2} = 0$

Step 4

Divide each term in $7 {(2 x - 1)}^{2} = 0$ by $7$ and simplify.

Divide each term in $7 {(2 x - 1)}^{2} = 0$ by $7$ .

$\frac{7 {(2 x - 1)}^{2}}{7} = \frac{0}{7}$

Simplify the left side.

Cancel the common factor of $7$ .

Cancel the common factor.

$\frac{7 {(2 x - 1)}^{2}}{7} = \frac{0}{7}$

Divide ${(2 x - 1)}^{2}$ by $1$ .

${(2 x - 1)}^{2} = \frac{0}{7}$

Simplify the right side.

Divide $0$ by $7$ .

${(2 x - 1)}^{2} = 0$

Step 5

Set the $2 x - 1$ equal to $0$ .

$2 x - 1 = 0$

Step 6

Solve for $x$ .

Add $1$ to both sides of the equation.

$2 x = 1$

Divide each term in $2 x = 1$ by $2$ and simplify.

Divide each term in $2 x = 1$ by $2$ .

$\frac{2 x}{2} = \frac{1}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{1}{2}$

Divide $x$ by $1$ .

$x = \frac{1}{2}$

Solve by Factoring 28x^2+7=28x

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