Solve (9-x)(x-4)=3x

$$( 9 - x ) ( x - 4 ) = 3 x$$

Solve for x (complex solution)

$$x=-\sqrt{11}i+5\approx 5-3.31662479i$$

$$x=5+\sqrt{11}i\approx 5+3.31662479i$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 9-x by x-4 and combine like terms.

$$13x-36-x^{2}=3x$$

Subtract 3x from both sides.

$$13x-36-x^{2}-3x=0$$

Combine 13x and -3x to get 10x.

$$10x-36-x^{2}=0$$

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.

$$-x^{2}+10x-36=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 10 for b, and -36 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-10±\sqrt{10^{2}-4\left(-1\right)\left(-36\right)}}{2\left(-1\right)}$$

Square 10.

$$x=\frac{-10±\sqrt{100-4\left(-1\right)\left(-36\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-10±\sqrt{100+4\left(-36\right)}}{2\left(-1\right)}$$

Multiply 4 times -36.

$$x=\frac{-10±\sqrt{100-144}}{2\left(-1\right)}$$

Add 100 to -144.

$$x=\frac{-10±\sqrt{-44}}{2\left(-1\right)}$$

Take the square root of -44.

$$x=\frac{-10±2\sqrt{11}i}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-10±2\sqrt{11}i}{-2}$$

Now solve the equation x=\frac{-10±2\sqrt{11}i}{-2} when ± is plus. Add -10 to 2i\sqrt{11}.

$$x=\frac{-10+2\sqrt{11}i}{-2}$$

Divide -10+2i\sqrt{11} by -2.

$$x=-\sqrt{11}i+5$$

Now solve the equation x=\frac{-10±2\sqrt{11}i}{-2} when ± is minus. Subtract 2i\sqrt{11} from -10.

$$x=\frac{-2\sqrt{11}i-10}{-2}$$

Divide -10-2i\sqrt{11} by -2.

$$x=5+\sqrt{11}i$$

The equation is now solved.

$$x=-\sqrt{11}i+5 x=5+\sqrt{11}i$$

Steps for Completing the Square

Use the distributive property to multiply 9-x by x-4 and combine like terms.

$$13x-36-x^{2}=3x$$

Subtract 3x from both sides.

$$13x-36-x^{2}-3x=0$$

Combine 13x and -3x to get 10x.

$$10x-36-x^{2}=0$$

Add 36 to both sides. Anything plus zero gives itself.

$$10x-x^{2}=36$$

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.

$$-x^{2}+10x=36$$

Divide both sides by -1.

$$\frac{-x^{2}+10x}{-1}=\frac{36}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{10}{-1}x=\frac{36}{-1}$$

Divide 10 by -1.

$$x^{2}-10x=\frac{36}{-1}$$

Divide 36 by -1.

$$x^{2}-10x=-36$$

Divide -10, the coefficient of the x term, by 2 to get -5. Then add the square of -5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-10x+\left(-5\right)^{2}=-36+\left(-5\right)^{2}$$

Square -5.

$$x^{2}-10x+25=-36+25$$

Add -36 to 25.

$$x^{2}-10x+25=-11$$

Factor x^{2}-10x+25. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-5\right)^{2}=-11$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-5\right)^{2}}=\sqrt{-11}$$

Simplify.

$$x-5=\sqrt{11}i x-5=-\sqrt{11}i$$

Add 5 to both sides of the equation.

$$x=5+\sqrt{11}i x=-\sqrt{11}i+5$$