Solve (x/2)^2-x=3

$$( \frac { x } { 2 } ) ^ { 2 } - x = 3$$

Solve for x

$$x=-2$$

$$x=6$$

Steps Using the Quadratic Formula

To raise \frac{x}{2} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{x^{2}}{2^{2}}-x=3$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x times \frac{2^{2}}{2^{2}}.

$$\frac{x^{2}}{2^{2}}-\frac{x\times 2^{2}}{2^{2}}=3$$

Since \frac{x^{2}}{2^{2}} and \frac{x\times 2^{2}}{2^{2}} have the same denominator, subtract them by subtracting their numerators.

$$\frac{x^{2}-x\times 2^{2}}{2^{2}}=3$$

Do the multiplications in x^{2}-x\times 2^{2}.

$$\frac{x^{2}-4x}{2^{2}}=3$$

Calculate 2 to the power of 2 and get 4.

$$\frac{x^{2}-4x}{4}=3$$

Divide each term of x^{2}-4x by 4 to get \frac{1}{4}x^{2}-x.

$$\frac{1}{4}x^{2}-x=3$$

Subtract 3 from both sides.

$$\frac{1}{4}x^{2}-x-3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{1}{4} for a, -1 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\times \frac{1}{4}\left(-3\right)}}{2\times \frac{1}{4}}$$

Multiply -4 times \frac{1}{4}.

$$x=\frac{-\left(-1\right)±\sqrt{1-\left(-3\right)}}{2\times \frac{1}{4}}$$

Multiply -1 times -3.

$$x=\frac{-\left(-1\right)±\sqrt{1+3}}{2\times \frac{1}{4}}$$

Add 1 to 3.

$$x=\frac{-\left(-1\right)±\sqrt{4}}{2\times \frac{1}{4}}$$

Take the square root of 4.

$$x=\frac{-\left(-1\right)±2}{2\times \frac{1}{4}}$$

The opposite of -1 is 1.

$$x=\frac{1±2}{2\times \frac{1}{4}}$$

Multiply 2 times \frac{1}{4}.

$$x=\frac{1±2}{\frac{1}{2}}$$

Now solve the equation x=\frac{1±2}{\frac{1}{2}} when ± is plus. Add 1 to 2.

$$x=\frac{3}{\frac{1}{2}}$$

Divide 3 by \frac{1}{2} by multiplying 3 by the reciprocal of \frac{1}{2}.

$$x=6$$

Now solve the equation x=\frac{1±2}{\frac{1}{2}} when ± is minus. Subtract 2 from 1.

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

Divide -1 by \frac{1}{2} by multiplying -1 by the reciprocal of \frac{1}{2}.

$$x=-2$$

The equation is now solved.

$$x=6 x=-2$$

Steps for Completing the Square

To raise \frac{x}{2} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{x^{2}}{2^{2}}-x=3$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x times \frac{2^{2}}{2^{2}}.

$$\frac{x^{2}}{2^{2}}-\frac{x\times 2^{2}}{2^{2}}=3$$

Since \frac{x^{2}}{2^{2}} and \frac{x\times 2^{2}}{2^{2}} have the same denominator, subtract them by subtracting their numerators.

$$\frac{x^{2}-x\times 2^{2}}{2^{2}}=3$$

Do the multiplications in x^{2}-x\times 2^{2}.

$$\frac{x^{2}-4x}{2^{2}}=3$$

Calculate 2 to the power of 2 and get 4.

$$\frac{x^{2}-4x}{4}=3$$

Divide each term of x^{2}-4x by 4 to get \frac{1}{4}x^{2}-x.

$$\frac{1}{4}x^{2}-x=3$$

Multiply both sides by 4.

$$\frac{\frac{1}{4}x^{2}-x}{\frac{1}{4}}=\frac{3}{\frac{1}{4}}$$

Dividing by \frac{1}{4} undoes the multiplication by \frac{1}{4}.

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

Divide -1 by \frac{1}{4} by multiplying -1 by the reciprocal of \frac{1}{4}.

$$x^{2}-4x=\frac{3}{\frac{1}{4}}$$

Divide 3 by \frac{1}{4} by multiplying 3 by the reciprocal of \frac{1}{4}.

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

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

$$x^{2}-4x+\left(-2\right)^{2}=12+\left(-2\right)^{2}$$

Square -2.

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

Add 12 to 4.

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

Factor x^{2}-4x+4. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-2\right)^{2}=16$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-2\right)^{2}}=\sqrt{16}$$

Simplify.

$$x-2=4 x-2=-4$$

Add 2 to both sides of the equation.

$$x=6 x=-2$$