Solve x^2+6x+{left(1/4x+6right)}^2-2left(1/4x+6right)=7

$${ x }^{ 2 } +6x+ { \left( \frac{ 1 }{ 4 } x+6 \right) }^{ 2 } -2 \left( \frac{ 1 }{ 4 } x+6 \right) = 7$$

Solve for x

$$x=-4$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\frac{1}{4}x+6\right)^{2}.

$$x^{2}+6x+\frac{1}{16}x^{2}+3x+36-2\left(\frac{1}{4}x+6\right)=7$$

Combine x^{2} and \frac{1}{16}x^{2} to get \frac{17}{16}x^{2}.

$$\frac{17}{16}x^{2}+6x+3x+36-2\left(\frac{1}{4}x+6\right)=7$$

Combine 6x and 3x to get 9x.

$$\frac{17}{16}x^{2}+9x+36-2\left(\frac{1}{4}x+6\right)=7$$

Use the distributive property to multiply -2 by \frac{1}{4}x+6.

$$\frac{17}{16}x^{2}+9x+36-\frac{1}{2}x-12=7$$

Combine 9x and -\frac{1}{2}x to get \frac{17}{2}x.

$$\frac{17}{16}x^{2}+\frac{17}{2}x+36-12=7$$

Subtract 12 from 36 to get 24.

$$\frac{17}{16}x^{2}+\frac{17}{2}x+24=7$$

Subtract 7 from both sides.

$$\frac{17}{16}x^{2}+\frac{17}{2}x+24-7=0$$

Subtract 7 from 24 to get 17.

$$\frac{17}{16}x^{2}+\frac{17}{2}x+17=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{17}{16} for a, \frac{17}{2} for b, and 17 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\frac{17}{2}±\sqrt{\left(\frac{17}{2}\right)^{2}-4\times \frac{17}{16}\times 17}}{2\times \frac{17}{16}}$$

Square \frac{17}{2} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\frac{17}{2}±\sqrt{\frac{289}{4}-4\times \frac{17}{16}\times 17}}{2\times \frac{17}{16}}$$

Multiply -4 times \frac{17}{16}.

$$x=\frac{-\frac{17}{2}±\sqrt{\frac{289}{4}-\frac{17}{4}\times 17}}{2\times \frac{17}{16}}$$

Multiply -\frac{17}{4} times 17.

$$x=\frac{-\frac{17}{2}±\sqrt{\frac{289-289}{4}}}{2\times \frac{17}{16}}$$

Add \frac{289}{4} to -\frac{289}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\frac{17}{2}±\sqrt{0}}{2\times \frac{17}{16}}$$

Take the square root of 0.

$$x=-\frac{\frac{17}{2}}{2\times \frac{17}{16}}$$

Multiply 2 times \frac{17}{16}.

$$x=-\frac{\frac{17}{2}}{\frac{17}{8}}$$

Divide -\frac{17}{2} by \frac{17}{8} by multiplying -\frac{17}{2} by the reciprocal of \frac{17}{8}.

$$x=-4$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\frac{1}{4}x+6\right)^{2}.

$$x^{2}+6x+\frac{1}{16}x^{2}+3x+36-2\left(\frac{1}{4}x+6\right)=7$$

Combine x^{2} and \frac{1}{16}x^{2} to get \frac{17}{16}x^{2}.

$$\frac{17}{16}x^{2}+6x+3x+36-2\left(\frac{1}{4}x+6\right)=7$$

Combine 6x and 3x to get 9x.

$$\frac{17}{16}x^{2}+9x+36-2\left(\frac{1}{4}x+6\right)=7$$

Use the distributive property to multiply -2 by \frac{1}{4}x+6.

$$\frac{17}{16}x^{2}+9x+36-\frac{1}{2}x-12=7$$

Combine 9x and -\frac{1}{2}x to get \frac{17}{2}x.

$$\frac{17}{16}x^{2}+\frac{17}{2}x+36-12=7$$

Subtract 12 from 36 to get 24.

$$\frac{17}{16}x^{2}+\frac{17}{2}x+24=7$$

Subtract 24 from both sides.

$$\frac{17}{16}x^{2}+\frac{17}{2}x=7-24$$

Subtract 24 from 7 to get -17.

$$\frac{17}{16}x^{2}+\frac{17}{2}x=-17$$

Divide both sides of the equation by \frac{17}{16}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{17}{16}x^{2}+\frac{17}{2}x}{\frac{17}{16}}=-\frac{17}{\frac{17}{16}}$$

Dividing by \frac{17}{16} undoes the multiplication by \frac{17}{16}.

$$x^{2}+\frac{\frac{17}{2}}{\frac{17}{16}}x=-\frac{17}{\frac{17}{16}}$$

Divide \frac{17}{2} by \frac{17}{16} by multiplying \frac{17}{2} by the reciprocal of \frac{17}{16}.

$$x^{2}+8x=-\frac{17}{\frac{17}{16}}$$

Divide -17 by \frac{17}{16} by multiplying -17 by the reciprocal of \frac{17}{16}.

$$x^{2}+8x=-16$$

Divide 8, the coefficient of the x term, by 2 to get 4. Then add the square of 4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+8x+4^{2}=-16+4^{2}$$

Square 4.

$$x^{2}+8x+16=-16+16$$

Add -16 to 16.

$$x^{2}+8x+16=0$$

Factor x^{2}+8x+16. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+4\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+4\right)^{2}}=\sqrt{0}$$

Simplify.

$$x+4=0 x+4=0$$

Subtract 4 from both sides of the equation.

$$x=-4 x=-4$$

The equation is now solved. Solutions are the same.

$$x=-4$$