Solve 3(2x-3)-3(3+2x^2)=0

$$3(2x-3)-3(3+2 { x }^{ 2 } )=0$$

Solve for x (complex solution)

$$x=\frac{1+\sqrt{11}i}{2}\approx 0.5+1.658312395i$$

$$x=\frac{-\sqrt{11}i+1}{2}\approx 0.5-1.658312395i$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 3 by 2x-3.

$$6x-9-3\left(3+2x^{2}\right)=0$$

Use the distributive property to multiply -3 by 3+2x^{2}.

$$6x-9-9-6x^{2}=0$$

Subtract 9 from -9 to get -18.

$$6x-18-6x^{2}=0$$

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.

$$-6x^{2}+6x-18=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -6 for a, 6 for b, and -18 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}-4\left(-6\right)\left(-18\right)}}{2\left(-6\right)}$$

Square 6.

$$x=\frac{-6±\sqrt{36-4\left(-6\right)\left(-18\right)}}{2\left(-6\right)}$$

Multiply -4 times -6.

$$x=\frac{-6±\sqrt{36+24\left(-18\right)}}{2\left(-6\right)}$$

Multiply 24 times -18.

$$x=\frac{-6±\sqrt{36-432}}{2\left(-6\right)}$$

Add 36 to -432.

$$x=\frac{-6±\sqrt{-396}}{2\left(-6\right)}$$

Take the square root of -396.

$$x=\frac{-6±6\sqrt{11}i}{2\left(-6\right)}$$

Multiply 2 times -6.

$$x=\frac{-6±6\sqrt{11}i}{-12}$$

Now solve the equation x=\frac{-6±6\sqrt{11}i}{-12} when ± is plus. Add -6 to 6i\sqrt{11}.

$$x=\frac{-6+6\sqrt{11}i}{-12}$$

Divide -6+6i\sqrt{11} by -12.

$$x=\frac{-\sqrt{11}i+1}{2}$$

Now solve the equation x=\frac{-6±6\sqrt{11}i}{-12} when ± is minus. Subtract 6i\sqrt{11} from -6.

$$x=\frac{-6\sqrt{11}i-6}{-12}$$

Divide -6-6i\sqrt{11} by -12.

$$x=\frac{1+\sqrt{11}i}{2}$$

The equation is now solved.

$$x=\frac{-\sqrt{11}i+1}{2} x=\frac{1+\sqrt{11}i}{2}$$

Steps for Completing the Square

Use the distributive property to multiply 3 by 2x-3.

$$6x-9-3\left(3+2x^{2}\right)=0$$

Use the distributive property to multiply -3 by 3+2x^{2}.

$$6x-9-9-6x^{2}=0$$

Subtract 9 from -9 to get -18.

$$6x-18-6x^{2}=0$$

Add 18 to both sides. Anything plus zero gives itself.

$$6x-6x^{2}=18$$

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.

$$-6x^{2}+6x=18$$

Divide both sides by -6.

$$\frac{-6x^{2}+6x}{-6}=\frac{18}{-6}$$

Dividing by -6 undoes the multiplication by -6.

$$x^{2}+\frac{6}{-6}x=\frac{18}{-6}$$

Divide 6 by -6.

$$x^{2}-x=\frac{18}{-6}$$

Divide 18 by -6.

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

Divide -1, the coefficient of the x term, by 2 to get -\frac{1}{2}. Then add the square of -\frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-x+\frac{1}{4}=-3+\frac{1}{4}$$

Add -3 to \frac{1}{4}.

$$x^{2}-x+\frac{1}{4}=-\frac{11}{4}$$

Factor x^{2}-x+\frac{1}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{2}\right)^{2}=-\frac{11}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{-\frac{11}{4}}$$

Simplify.

$$x-\frac{1}{2}=\frac{\sqrt{11}i}{2} x-\frac{1}{2}=-\frac{\sqrt{11}i}{2}$$

Add \frac{1}{2} to both sides of the equation.

$$x=\frac{1+\sqrt{11}i}{2} x=\frac{-\sqrt{11}i+1}{2}$$