Solve (3x-2)^2+5x^2=(3x+2)(3x-2)

$$( 3 x - 2 ) ^ { 2 } + 5 x ^ { 2 } = ( 3 x + 2 ) ( 3 x - 2 )$$

Solve for x (complex solution)

$$x=\frac{6}{5}+\frac{2}{5}i=1.2+0.4i$$

$$x=\frac{6}{5}-\frac{2}{5}i=1.2-0.4i$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-2\right)^{2}.

$$9x^{2}-12x+4+5x^{2}=\left(3x+2\right)\left(3x-2\right)$$

Combine 9x^{2} and 5x^{2} to get 14x^{2}.

$$14x^{2}-12x+4=\left(3x+2\right)\left(3x-2\right)$$

Consider \left(3x+2\right)\left(3x-2\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 2.

$$14x^{2}-12x+4=\left(3x\right)^{2}-4$$

Expand \left(3x\right)^{2}.

$$14x^{2}-12x+4=3^{2}x^{2}-4$$

Calculate 3 to the power of 2 and get 9.

$$14x^{2}-12x+4=9x^{2}-4$$

Subtract 9x^{2} from both sides.

$$14x^{2}-12x+4-9x^{2}=-4$$

Combine 14x^{2} and -9x^{2} to get 5x^{2}.

$$5x^{2}-12x+4=-4$$

Add 4 to both sides.

$$5x^{2}-12x+4+4=0$$

Add 4 and 4 to get 8.

$$5x^{2}-12x+8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -12 for b, and 8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 5\times 8}}{2\times 5}$$

Square -12.

$$x=\frac{-\left(-12\right)±\sqrt{144-4\times 5\times 8}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-\left(-12\right)±\sqrt{144-20\times 8}}{2\times 5}$$

Multiply -20 times 8.

$$x=\frac{-\left(-12\right)±\sqrt{144-160}}{2\times 5}$$

Add 144 to -160.

$$x=\frac{-\left(-12\right)±\sqrt{-16}}{2\times 5}$$

Take the square root of -16.

$$x=\frac{-\left(-12\right)±4i}{2\times 5}$$

The opposite of -12 is 12.

$$x=\frac{12±4i}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{12±4i}{10}$$

Now solve the equation x=\frac{12±4i}{10} when ± is plus. Add 12 to 4i.

$$x=\frac{12+4i}{10}$$

Divide 12+4i by 10.

$$x=\frac{6}{5}+\frac{2}{5}i$$

Now solve the equation x=\frac{12±4i}{10} when ± is minus. Subtract 4i from 12.

$$x=\frac{12-4i}{10}$$

Divide 12-4i by 10.

$$x=\frac{6}{5}-\frac{2}{5}i$$

The equation is now solved.

$$x=\frac{6}{5}+\frac{2}{5}i x=\frac{6}{5}-\frac{2}{5}i$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-2\right)^{2}.

$$9x^{2}-12x+4+5x^{2}=\left(3x+2\right)\left(3x-2\right)$$

Combine 9x^{2} and 5x^{2} to get 14x^{2}.

$$14x^{2}-12x+4=\left(3x+2\right)\left(3x-2\right)$$

Consider \left(3x+2\right)\left(3x-2\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 2.

$$14x^{2}-12x+4=\left(3x\right)^{2}-4$$

Expand \left(3x\right)^{2}.

$$14x^{2}-12x+4=3^{2}x^{2}-4$$

Calculate 3 to the power of 2 and get 9.

$$14x^{2}-12x+4=9x^{2}-4$$

Subtract 9x^{2} from both sides.

$$14x^{2}-12x+4-9x^{2}=-4$$

Combine 14x^{2} and -9x^{2} to get 5x^{2}.

$$5x^{2}-12x+4=-4$$

Subtract 4 from both sides.

$$5x^{2}-12x=-4-4$$

Subtract 4 from -4 to get -8.

$$5x^{2}-12x=-8$$

Divide both sides by 5.

$$\frac{5x^{2}-12x}{5}=-\frac{8}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}-\frac{12}{5}x=-\frac{8}{5}$$

Divide -\frac{12}{5}, the coefficient of the x term, by 2 to get -\frac{6}{5}. Then add the square of -\frac{6}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{12}{5}x+\left(-\frac{6}{5}\right)^{2}=-\frac{8}{5}+\left(-\frac{6}{5}\right)^{2}$$

Square -\frac{6}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{12}{5}x+\frac{36}{25}=-\frac{8}{5}+\frac{36}{25}$$

Add -\frac{8}{5} to \frac{36}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{12}{5}x+\frac{36}{25}=-\frac{4}{25}$$

Factor x^{2}-\frac{12}{5}x+\frac{36}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{6}{5}\right)^{2}=-\frac{4}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{6}{5}\right)^{2}}=\sqrt{-\frac{4}{25}}$$

Simplify.

$$x-\frac{6}{5}=\frac{2}{5}i x-\frac{6}{5}=-\frac{2}{5}i$$

Add \frac{6}{5} to both sides of the equation.

$$x=\frac{6}{5}+\frac{2}{5}i x=\frac{6}{5}-\frac{2}{5}i$$