Solve [(7-x)x/2]/2=3

$$\frac { [ ( 7 - x ) \frac { x } { 2 } ] } { 2 } = 3$$

Solve for x

$$x=3$$

$$x=4$$

Steps Using the Quadratic Formula

Multiply both sides by 2.

$$\left(7-x\right)\times \frac{x}{2}=3\times 2$$

Multiply both sides of the equation by 2.

$$\left(7-x\right)x=6\times 2$$

Use the distributive property to multiply 7-x by x.

$$7x-x^{2}=6\times 2$$

Multiply 6 and 2 to get 12.

$$7x-x^{2}=12$$

Subtract 12 from both sides.

$$7x-x^{2}-12=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-x^{2}+7x-12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 7 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-7±\sqrt{7^{2}-4\left(-1\right)\left(-12\right)}}{2\left(-1\right)}$$

Square 7.

$$x=\frac{-7±\sqrt{49-4\left(-1\right)\left(-12\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-7±\sqrt{49+4\left(-12\right)}}{2\left(-1\right)}$$

Multiply 4 times -12.

$$x=\frac{-7±\sqrt{49-48}}{2\left(-1\right)}$$

Add 49 to -48.

$$x=\frac{-7±\sqrt{1}}{2\left(-1\right)}$$

Take the square root of 1.

$$x=\frac{-7±1}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-7±1}{-2}$$

Now solve the equation x=\frac{-7±1}{-2} when ± is plus. Add -7 to 1.

$$x=-\frac{6}{-2}$$

Divide -6 by -2.

$$x=3$$

Now solve the equation x=\frac{-7±1}{-2} when ± is minus. Subtract 1 from -7.

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

Divide -8 by -2.

$$x=4$$

The equation is now solved.

$$x=3 x=4$$

Steps for Completing the Square

Multiply both sides by 2.

$$\left(7-x\right)\times \frac{x}{2}=3\times 2$$

Multiply both sides of the equation by 2.

$$\left(7-x\right)x=6\times 2$$

Use the distributive property to multiply 7-x by x.

$$7x-x^{2}=6\times 2$$

Multiply 6 and 2 to get 12.

$$7x-x^{2}=12$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-x^{2}+7x=12$$

Divide both sides by -1.

$$\frac{-x^{2}+7x}{-1}=\frac{12}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{7}{-1}x=\frac{12}{-1}$$

Divide 7 by -1.

$$x^{2}-7x=\frac{12}{-1}$$

Divide 12 by -1.

$$x^{2}-7x=-12$$

Divide -7, the coefficient of the x term, by 2 to get -\frac{7}{2}. Then add the square of -\frac{7}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-7x+\left(-\frac{7}{2}\right)^{2}=-12+\left(-\frac{7}{2}\right)^{2}$$

Square -\frac{7}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-7x+\frac{49}{4}=-12+\frac{49}{4}$$

Add -12 to \frac{49}{4}.

$$x^{2}-7x+\frac{49}{4}=\frac{1}{4}$$

Factor x^{2}-7x+\frac{49}{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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-\frac{7}{2}=\frac{1}{2} x-\frac{7}{2}=-\frac{1}{2}$$

Add \frac{7}{2} to both sides of the equation.

$$x=4 x=3$$