Solve 2x(12-2x)=40

$$2x(12-2x)=40$$

Solve for x (complex solution)

$$x=3+i$$

$$x=3-i$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 2x by 12-2x.

$$24x-4x^{2}=40$$

Subtract 40 from both sides.

$$24x-4x^{2}-40=0$$

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.

$$-4x^{2}+24x-40=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, 24 for b, and -40 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-24±\sqrt{24^{2}-4\left(-4\right)\left(-40\right)}}{2\left(-4\right)}$$

Square 24.

$$x=\frac{-24±\sqrt{576-4\left(-4\right)\left(-40\right)}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$x=\frac{-24±\sqrt{576+16\left(-40\right)}}{2\left(-4\right)}$$

Multiply 16 times -40.

$$x=\frac{-24±\sqrt{576-640}}{2\left(-4\right)}$$

Add 576 to -640.

$$x=\frac{-24±\sqrt{-64}}{2\left(-4\right)}$$

Take the square root of -64.

$$x=\frac{-24±8i}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{-24±8i}{-8}$$

Now solve the equation x=\frac{-24±8i}{-8} when ± is plus. Add -24 to 8i.

$$x=\frac{-24+8i}{-8}$$

Divide -24+8i by -8.

$$x=3-i$$

Now solve the equation x=\frac{-24±8i}{-8} when ± is minus. Subtract 8i from -24.

$$x=\frac{-24-8i}{-8}$$

Divide -24-8i by -8.

$$x=3+i$$

The equation is now solved.

$$x=3-i x=3+i$$

Steps for Completing the Square

Use the distributive property to multiply 2x by 12-2x.

$$24x-4x^{2}=40$$

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.

$$-4x^{2}+24x=40$$

Divide both sides by -4.

$$\frac{-4x^{2}+24x}{-4}=\frac{40}{-4}$$

Dividing by -4 undoes the multiplication by -4.

$$x^{2}+\frac{24}{-4}x=\frac{40}{-4}$$

Divide 24 by -4.

$$x^{2}-6x=\frac{40}{-4}$$

Divide 40 by -4.

$$x^{2}-6x=-10$$

Divide -6, the coefficient of the x term, by 2 to get -3. Then add the square of -3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-6x+\left(-3\right)^{2}=-10+\left(-3\right)^{2}$$

Square -3.

$$x^{2}-6x+9=-10+9$$

Add -10 to 9.

$$x^{2}-6x+9=-1$$

Factor x^{2}-6x+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

$$\sqrt{\left(x-3\right)^{2}}=\sqrt{-1}$$

Simplify.

$$x-3=i x-3=-i$$

Add 3 to both sides of the equation.

$$x=3+i x=3-i$$