Solve (18-3x)x=40

$$(18-3x)x=40$$

Solve for x (complex solution)

$$x=-\frac{\sqrt{39}i}{3}+3\approx 3-2.081665999i$$

$$x=\frac{\sqrt{39}i}{3}+3\approx 3+2.081665999i$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 18-3x by x.

$$18x-3x^{2}=40$$

Subtract 40 from both sides.

$$18x-3x^{2}-40=0$$

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.

$$-3x^{2}+18x-40=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, 18 for b, and -40 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-18±\sqrt{18^{2}-4\left(-3\right)\left(-40\right)}}{2\left(-3\right)}$$

Square 18.

$$x=\frac{-18±\sqrt{324-4\left(-3\right)\left(-40\right)}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-18±\sqrt{324+12\left(-40\right)}}{2\left(-3\right)}$$

Multiply 12 times -40.

$$x=\frac{-18±\sqrt{324-480}}{2\left(-3\right)}$$

Add 324 to -480.

$$x=\frac{-18±\sqrt{-156}}{2\left(-3\right)}$$

Take the square root of -156.

$$x=\frac{-18±2\sqrt{39}i}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{-18±2\sqrt{39}i}{-6}$$

Now solve the equation x=\frac{-18±2\sqrt{39}i}{-6} when ± is plus. Add -18 to 2i\sqrt{39}.

$$x=\frac{-18+2\sqrt{39}i}{-6}$$

Divide -18+2i\sqrt{39} by -6.

$$x=-\frac{\sqrt{39}i}{3}+3$$

Now solve the equation x=\frac{-18±2\sqrt{39}i}{-6} when ± is minus. Subtract 2i\sqrt{39} from -18.

$$x=\frac{-2\sqrt{39}i-18}{-6}$$

Divide -18-2i\sqrt{39} by -6.

$$x=\frac{\sqrt{39}i}{3}+3$$

The equation is now solved.

$$x=-\frac{\sqrt{39}i}{3}+3 x=\frac{\sqrt{39}i}{3}+3$$

Steps for Completing the Square

Use the distributive property to multiply 18-3x by x.

$$18x-3x^{2}=40$$

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.

$$-3x^{2}+18x=40$$

Divide both sides by -3.

$$\frac{-3x^{2}+18x}{-3}=\frac{40}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$x^{2}+\frac{18}{-3}x=\frac{40}{-3}$$

Divide 18 by -3.

$$x^{2}-6x=\frac{40}{-3}$$

Divide 40 by -3.

$$x^{2}-6x=-\frac{40}{3}$$

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.

$$x^{2}-6x+\left(-3\right)^{2}=-\frac{40}{3}+\left(-3\right)^{2}$$

Square -3.

$$x^{2}-6x+9=-\frac{40}{3}+9$$

Add -\frac{40}{3} to 9.

$$x^{2}-6x+9=-\frac{13}{3}$$

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}.

$$\left(x-3\right)^{2}=-\frac{13}{3}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-3\right)^{2}}=\sqrt{-\frac{13}{3}}$$

Simplify.

$$x-3=\frac{\sqrt{39}i}{3} x-3=-\frac{\sqrt{39}i}{3}$$

Add 3 to both sides of the equation.

$$x=\frac{\sqrt{39}i}{3}+3 x=-\frac{\sqrt{39}i}{3}+3$$