Solve by Completing the Square 4x^2-30x=12+10x

$4 x^{2} - 30 x = 12 + 10 x$

Simplify the equation into a proper form to complete the square.

Reorder $12$ and $10 x$ .

$4 x^{2} - 30 x = 10 x + 12$

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

Subtract $10 x$ from both sides of the equation.

$4 x^{2} - 30 x - 10 x = 12$

Subtract $10 x$ from $- 30 x$ .

$4 x^{2} - 40 x = 12$

Divide each term in $4 x^{2} - 40 x = 12$ by $4$ and simplify.

Divide each term in $4 x^{2} - 40 x = 12$ by $4$ .

$\frac{4 x^{2}}{4} + \frac{- 40 x}{4} = \frac{12}{4}$

Simplify the left side.

Simplify each term.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 x^{2}}{4} + \frac{- 40 x}{4} = \frac{12}{4}$

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

$x^{2} + \frac{- 40 x}{4} = \frac{12}{4}$

Cancel the common factor of $- 40$ and $4$ .

Factor $4$ out of $- 40 x$ .

$x^{2} + \frac{4 (- 10 x)}{4} = \frac{12}{4}$

Cancel the common factors.

Factor $4$ out of $4$ .

$x^{2} + \frac{4 (- 10 x)}{4 (1)} = \frac{12}{4}$

Cancel the common factor.

$x^{2} + \frac{4 (- 10 x)}{4 \cdot 1} = \frac{12}{4}$

Rewrite the expression.

$x^{2} + \frac{- 10 x}{1} = \frac{12}{4}$

Divide $- 10 x$ by $1$ .

$x^{2} - 10 x = \frac{12}{4}$

Simplify the right side.

Divide $12$ by $4$ .

$x^{2} - 10 x = 3$

To create a trinomial square on the left side of the equation, find a value that is equal to the square of half of $b$ .

${(\frac{b}{2})}^{2} = {(- 5)}^{2}$

Add the term to each side of the equation.

$x^{2} - 10 x + {(- 5)}^{2} = 3 + {(- 5)}^{2}$

Simplify the equation.

Simplify the left side.

Raise $- 5$ to the power of $2$ .

$x^{2} - 10 x + 25 = 3 + {(- 5)}^{2}$

Simplify the right side.

Simplify $3 + {(- 5)}^{2}$ .

Raise $- 5$ to the power of $2$ .

$x^{2} - 10 x + 25 = 3 + 25$

Add $3$ and $25$ .

$x^{2} - 10 x + 25 = 28$

Factor the perfect trinomial square into ${(x - 5)}^{2}$ .

${(x - 5)}^{2} = 28$

Solve the equation for $x$ .

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

$x - 5 = \pm \sqrt{28}$

Simplify $\pm \sqrt{28}$ .

Rewrite $28$ as $2^{2} \cdot 7$ .

Factor $4$ out of $28$ .

$x - 5 = \pm \sqrt{4 (7)}$

Rewrite $4$ as $2^{2}$ .

$x - 5 = \pm \sqrt{2^{2} \cdot 7}$

Pull terms out from under the radical.

$x - 5 = \pm 2 \sqrt{7}$

Add $5$ to both sides of the equation.

$x = \pm 2 \sqrt{7} + 5$

The result can be shown in multiple forms.

Exact Form:

$x = \pm 2 \sqrt{7} + 5$

Decimal Form:

$x = 10.29150262 \dots, - 0.29150262 \dots$

Solve by Completing the Square 4x^2-30x=12+10x

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