Solve x+3*(x-3)-(x^2-4x)=9

$$x + 3 \cdot ( x - 3 ) - ( x ^ { 2 } - 4 x ) = 9$$

Solve for x (complex solution)

$$x=-\sqrt{2}i+4\approx 4-1.414213562i$$

$$x=4+\sqrt{2}i\approx 4+1.414213562i$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 3 by x-3.

$$x+3x-9-\left(x^{2}-4x\right)=9$$

Combine x and 3x to get 4x.

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

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

$$4x-9-x^{2}+4x=9$$

Combine 4x and 4x to get 8x.

$$8x-9-x^{2}=9$$

Subtract 9 from both sides.

$$8x-9-x^{2}-9=0$$

Subtract 9 from -9 to get -18.

$$8x-18-x^{2}=0$$

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.

$$-x^{2}+8x-18=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 8 for b, and -18 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-8±\sqrt{8^{2}-4\left(-1\right)\left(-18\right)}}{2\left(-1\right)}$$

Square 8.

$$x=\frac{-8±\sqrt{64-4\left(-1\right)\left(-18\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-8±\sqrt{64+4\left(-18\right)}}{2\left(-1\right)}$$

Multiply 4 times -18.

$$x=\frac{-8±\sqrt{64-72}}{2\left(-1\right)}$$

Add 64 to -72.

$$x=\frac{-8±\sqrt{-8}}{2\left(-1\right)}$$

Take the square root of -8.

$$x=\frac{-8±2\sqrt{2}i}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-8±2\sqrt{2}i}{-2}$$

Now solve the equation x=\frac{-8±2\sqrt{2}i}{-2} when ± is plus. Add -8 to 2i\sqrt{2}.

$$x=\frac{-8+2\sqrt{2}i}{-2}$$

Divide -8+2i\sqrt{2} by -2.

$$x=-\sqrt{2}i+4$$

Now solve the equation x=\frac{-8±2\sqrt{2}i}{-2} when ± is minus. Subtract 2i\sqrt{2} from -8.

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

Divide -8-2i\sqrt{2} by -2.

$$x=4+\sqrt{2}i$$

The equation is now solved.

$$x=-\sqrt{2}i+4 x=4+\sqrt{2}i$$

Steps for Completing the Square

Use the distributive property to multiply 3 by x-3.

$$x+3x-9-\left(x^{2}-4x\right)=9$$

Combine x and 3x to get 4x.

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

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

$$4x-9-x^{2}+4x=9$$

Combine 4x and 4x to get 8x.

$$8x-9-x^{2}=9$$

Add 9 to both sides.

$$8x-x^{2}=9+9$$

Add 9 and 9 to get 18.

$$8x-x^{2}=18$$

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.

$$-x^{2}+8x=18$$

Divide both sides by -1.

$$\frac{-x^{2}+8x}{-1}=\frac{18}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{8}{-1}x=\frac{18}{-1}$$

Divide 8 by -1.

$$x^{2}-8x=\frac{18}{-1}$$

Divide 18 by -1.

$$x^{2}-8x=-18$$

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.

$$x^{2}-8x+\left(-4\right)^{2}=-18+\left(-4\right)^{2}$$

Square -4.

$$x^{2}-8x+16=-18+16$$

Add -18 to 16.

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

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-4=\sqrt{2}i x-4=-\sqrt{2}i$$

Add 4 to both sides of the equation.

$$x=4+\sqrt{2}i x=-\sqrt{2}i+4$$