Solve 3x^2-18x=-135

$$3 x ^ { 2 } - 18 x = - 135$$

Solve for x (complex solution)

$$x=3+6i$$

$$x=3-6i$$

Steps Using the Quadratic Formula

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.

$$3x^{2}-18x=-135$$

Add 135 to both sides of the equation.

$$3x^{2}-18x-\left(-135\right)=-135-\left(-135\right)$$

Subtracting -135 from itself leaves 0.

$$3x^{2}-18x-\left(-135\right)=0$$

Subtract -135 from 0.

$$3x^{2}-18x+135=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -18 for b, and 135 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-18\right)±\sqrt{\left(-18\right)^{2}-4\times 3\times 135}}{2\times 3}$$

Square -18.

$$x=\frac{-\left(-18\right)±\sqrt{324-4\times 3\times 135}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-18\right)±\sqrt{324-12\times 135}}{2\times 3}$$

Multiply -12 times 135.

$$x=\frac{-\left(-18\right)±\sqrt{324-1620}}{2\times 3}$$

Add 324 to -1620.

$$x=\frac{-\left(-18\right)±\sqrt{-1296}}{2\times 3}$$

Take the square root of -1296.

$$x=\frac{-\left(-18\right)±36i}{2\times 3}$$

The opposite of -18 is 18.

$$x=\frac{18±36i}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{18±36i}{6}$$

Now solve the equation x=\frac{18±36i}{6} when ± is plus. Add 18 to 36i.

$$x=\frac{18+36i}{6}$$

Divide 18+36i by 6.

$$x=3+6i$$

Now solve the equation x=\frac{18±36i}{6} when ± is minus. Subtract 36i from 18.

$$x=\frac{18-36i}{6}$$

Divide 18-36i by 6.

$$x=3-6i$$

The equation is now solved.

$$x=3+6i x=3-6i$$

Steps for Completing the Square

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.

$$3x^{2}-18x=-135$$

Divide both sides by 3.

$$\frac{3x^{2}-18x}{3}=-\frac{135}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\left(-\frac{18}{3}\right)x=-\frac{135}{3}$$

Divide -18 by 3.

$$x^{2}-6x=-\frac{135}{3}$$

Divide -135 by 3.

$$x^{2}-6x=-45$$

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.

$$x^{2}-6x+\left(-3\right)^{2}=-45+\left(-3\right)^{2}$$

Square -3.

$$x^{2}-6x+9=-45+9$$

Add -45 to 9.

$$x^{2}-6x+9=-36$$

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}.

$$\left(x-3\right)^{2}=-36$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-3\right)^{2}}=\sqrt{-36}$$

Simplify.

$$x-3=6i x-3=-6i$$

Add 3 to both sides of the equation.

$$x=3+6i x=3-6i$$