Solve (x-3)(-x+3)-x(x-4)=0

$$( x - 3 ) ( - x + 3 ) - x ( x - 4 ) = 0$$

Solve for x

$$x = \frac{\sqrt{7} + 5}{2} \approx 3.822875656$$

$$x = \frac{5 - \sqrt{7}}{2} \approx 1.177124344$$

Steps Using the Quadratic Formula

Apply the distributive property by multiplying each term of x-3 by each term of -x+3.

$$x\left(-x\right)+3x-3\left(-x\right)-9-x\left(x-4\right)=0$$

Multiply -3 and -1 to get 3.

$$x\left(-x\right)+3x+3x-9-x\left(x-4\right)=0$$

Combine 3x and 3x to get 6x.

$$x\left(-x\right)+6x-9-x\left(x-4\right)=0$$

Use the distributive property to multiply x by x-4.

$$x\left(-x\right)+6x-9-\left(x^{2}-4x\right)=0$$

To find the opposite of x^{2}-4x, find the opposite of each term.

$$x\left(-x\right)+6x-9-x^{2}-\left(-4x\right)=0$$

The opposite of -4x is 4x.

$$x\left(-x\right)+6x-9-x^{2}+4x=0$$

Combine 6x and 4x to get 10x.

$$x\left(-x\right)+10x-9-x^{2}=0$$

Multiply x and x to get x^{2}.

$$x^{2}\left(-1\right)+10x-9-x^{2}=0$$

Combine x^{2}\left(-1\right) and -x^{2} to get -2x^{2}.

$$-2x^{2}+10x-9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, 10 for b, and -9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-10±\sqrt{10^{2}-4\left(-2\right)\left(-9\right)}}{2\left(-2\right)}$$

Square 10.

$$x=\frac{-10±\sqrt{100-4\left(-2\right)\left(-9\right)}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-10±\sqrt{100+8\left(-9\right)}}{2\left(-2\right)}$$

Multiply 8 times -9.

$$x=\frac{-10±\sqrt{100-72}}{2\left(-2\right)}$$

Add 100 to -72.

$$x=\frac{-10±\sqrt{28}}{2\left(-2\right)}$$

Take the square root of 28.

$$x=\frac{-10±2\sqrt{7}}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-10±2\sqrt{7}}{-4}$$

Now solve the equation x=\frac{-10±2\sqrt{7}}{-4} when ± is plus. Add -10 to 2\sqrt{7}.

$$x=\frac{2\sqrt{7}-10}{-4}$$

Divide -10+2\sqrt{7} by -4.

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

Now solve the equation x=\frac{-10±2\sqrt{7}}{-4} when ± is minus. Subtract 2\sqrt{7} from -10.

$$x=\frac{-2\sqrt{7}-10}{-4}$$

Divide -10-2\sqrt{7} by -4.

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

The equation is now solved.

$$x=\frac{5-\sqrt{7}}{2} x=\frac{\sqrt{7}+5}{2}$$

Steps for Completing the Square

Apply the distributive property by multiplying each term of x-3 by each term of -x+3.

$$x\left(-x\right)+3x-3\left(-x\right)-9-x\left(x-4\right)=0$$

Multiply -3 and -1 to get 3.

$$x\left(-x\right)+3x+3x-9-x\left(x-4\right)=0$$

Combine 3x and 3x to get 6x.

$$x\left(-x\right)+6x-9-x\left(x-4\right)=0$$

Use the distributive property to multiply x by x-4.

$$x\left(-x\right)+6x-9-\left(x^{2}-4x\right)=0$$

To find the opposite of x^{2}-4x, find the opposite of each term.

$$x\left(-x\right)+6x-9-x^{2}-\left(-4x\right)=0$$

The opposite of -4x is 4x.

$$x\left(-x\right)+6x-9-x^{2}+4x=0$$

Combine 6x and 4x to get 10x.

$$x\left(-x\right)+10x-9-x^{2}=0$$

Add 9 to both sides. Anything plus zero gives itself.

$$x\left(-x\right)+10x-x^{2}=9$$

Multiply x and x to get x^{2}.

$$x^{2}\left(-1\right)+10x-x^{2}=9$$

Combine x^{2}\left(-1\right) and -x^{2} to get -2x^{2}.

$$-2x^{2}+10x=9$$

Divide both sides by -2.

$$\frac{-2x^{2}+10x}{-2}=\frac{9}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\frac{10}{-2}x=\frac{9}{-2}$$

Divide 10 by -2.

$$x^{2}-5x=\frac{9}{-2}$$

Divide 9 by -2.

$$x^{2}-5x=-\frac{9}{2}$$

Divide -5, the coefficient of the x term, by 2 to get -\frac{5}{2}. Then add the square of -\frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-5x+\left(-\frac{5}{2}\right)^{2}=-\frac{9}{2}+\left(-\frac{5}{2}\right)^{2}$$

Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-5x+\frac{25}{4}=-\frac{9}{2}+\frac{25}{4}$$

Add -\frac{9}{2} to \frac{25}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-5x+\frac{25}{4}=\frac{7}{4}$$

Factor x^{2}-5x+\frac{25}{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}.

$$\left(x-\frac{5}{2}\right)^{2}=\frac{7}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{\frac{7}{4}}$$

Simplify.

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

Add \frac{5}{2} to both sides of the equation.

$$x=\frac{\sqrt{7}+5}{2} x=\frac{5-\sqrt{7}}{2}$$