Solve 3(x+x-30)=2x(x+6)

$$3 ( x + x - 30 ) = 2 x ( x + 6 )$$

Solve for x (complex solution)

$$x=\frac{-3+3\sqrt{19}i}{2}\approx -1.5+6.538348415i$$

$$x=\frac{-3\sqrt{19}i-3}{2}\approx -1.5-6.538348415i$$

Steps Using the Quadratic Formula

Combine x and x to get 2x.

$$3\left(2x-30\right)=2x\left(x+6\right)$$

Use the distributive property to multiply 3 by 2x-30.

$$6x-90=2x\left(x+6\right)$$

Use the distributive property to multiply 2x by x+6.

$$6x-90=2x^{2}+12x$$

Subtract 2x^{2} from both sides.

$$6x-90-2x^{2}=12x$$

Subtract 12x from both sides.

$$6x-90-2x^{2}-12x=0$$

Combine 6x and -12x to get -6x.

$$-6x-90-2x^{2}=0$$

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.

$$-2x^{2}-6x-90=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, -6 for b, and -90 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\left(-2\right)\left(-90\right)}}{2\left(-2\right)}$$

Square -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\left(-2\right)\left(-90\right)}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-\left(-6\right)±\sqrt{36+8\left(-90\right)}}{2\left(-2\right)}$$

Multiply 8 times -90.

$$x=\frac{-\left(-6\right)±\sqrt{36-720}}{2\left(-2\right)}$$

Add 36 to -720.

$$x=\frac{-\left(-6\right)±\sqrt{-684}}{2\left(-2\right)}$$

Take the square root of -684.

$$x=\frac{-\left(-6\right)±6\sqrt{19}i}{2\left(-2\right)}$$

The opposite of -6 is 6.

$$x=\frac{6±6\sqrt{19}i}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{6±6\sqrt{19}i}{-4}$$

Now solve the equation x=\frac{6±6\sqrt{19}i}{-4} when ± is plus. Add 6 to 6i\sqrt{19}.

$$x=\frac{6+6\sqrt{19}i}{-4}$$

Divide 6+6i\sqrt{19} by -4.

$$x=\frac{-3\sqrt{19}i-3}{2}$$

Now solve the equation x=\frac{6±6\sqrt{19}i}{-4} when ± is minus. Subtract 6i\sqrt{19} from 6.

$$x=\frac{-6\sqrt{19}i+6}{-4}$$

Divide 6-6i\sqrt{19} by -4.

$$x=\frac{-3+3\sqrt{19}i}{2}$$

The equation is now solved.

$$x=\frac{-3\sqrt{19}i-3}{2} x=\frac{-3+3\sqrt{19}i}{2}$$

Steps for Completing the Square

Combine x and x to get 2x.

$$3\left(2x-30\right)=2x\left(x+6\right)$$

Use the distributive property to multiply 3 by 2x-30.

$$6x-90=2x\left(x+6\right)$$

Use the distributive property to multiply 2x by x+6.

$$6x-90=2x^{2}+12x$$

Subtract 2x^{2} from both sides.

$$6x-90-2x^{2}=12x$$

Subtract 12x from both sides.

$$6x-90-2x^{2}-12x=0$$

Combine 6x and -12x to get -6x.

$$-6x-90-2x^{2}=0$$

Add 90 to both sides. Anything plus zero gives itself.

$$-6x-2x^{2}=90$$

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.

$$-2x^{2}-6x=90$$

Divide both sides by -2.

$$\frac{-2x^{2}-6x}{-2}=\frac{90}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\left(-\frac{6}{-2}\right)x=\frac{90}{-2}$$

Divide -6 by -2.

$$x^{2}+3x=\frac{90}{-2}$$

Divide 90 by -2.

$$x^{2}+3x=-45$$

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

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

Square \frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+3x+\frac{9}{4}=-45+\frac{9}{4}$$

Add -45 to \frac{9}{4}.

$$x^{2}+3x+\frac{9}{4}=-\frac{171}{4}$$

Factor x^{2}+3x+\frac{9}{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}.

$$\left(x+\frac{3}{2}\right)^{2}=-\frac{171}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{2}\right)^{2}}=\sqrt{-\frac{171}{4}}$$

Simplify.

$$x+\frac{3}{2}=\frac{3\sqrt{19}i}{2} x+\frac{3}{2}=-\frac{3\sqrt{19}i}{2}$$

Subtract \frac{3}{2} from both sides of the equation.

$$x=\frac{-3+3\sqrt{19}i}{2} x=\frac{-3\sqrt{19}i-3}{2}$$