Solve 7-2(3x-1)^2=47

$$7 - 2 ( 3 x - 1 ) ^ { 2 } = 47$$

Solve for x (complex solution)

$$x=\frac{-2\sqrt{5}i+1}{3}\approx 0.333333333-1.490711985i$$

$$x=\frac{1+2\sqrt{5}i}{3}\approx 0.333333333+1.490711985i$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-1\right)^{2}.

$$7-2\left(9x^{2}-6x+1\right)=47$$

Use the distributive property to multiply -2 by 9x^{2}-6x+1.

$$7-18x^{2}+12x-2=47$$

Subtract 2 from 7 to get 5.

$$5-18x^{2}+12x=47$$

Subtract 47 from both sides.

$$5-18x^{2}+12x-47=0$$

Subtract 47 from 5 to get -42.

$$-42-18x^{2}+12x=0$$

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.

$$-18x^{2}+12x-42=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -18 for a, 12 for b, and -42 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-12±\sqrt{12^{2}-4\left(-18\right)\left(-42\right)}}{2\left(-18\right)}$$

Square 12.

$$x=\frac{-12±\sqrt{144-4\left(-18\right)\left(-42\right)}}{2\left(-18\right)}$$

Multiply -4 times -18.

$$x=\frac{-12±\sqrt{144+72\left(-42\right)}}{2\left(-18\right)}$$

Multiply 72 times -42.

$$x=\frac{-12±\sqrt{144-3024}}{2\left(-18\right)}$$

Add 144 to -3024.

$$x=\frac{-12±\sqrt{-2880}}{2\left(-18\right)}$$

Take the square root of -2880.

$$x=\frac{-12±24\sqrt{5}i}{2\left(-18\right)}$$

Multiply 2 times -18.

$$x=\frac{-12±24\sqrt{5}i}{-36}$$

Now solve the equation x=\frac{-12±24\sqrt{5}i}{-36} when ± is plus. Add -12 to 24i\sqrt{5}.

$$x=\frac{-12+24\sqrt{5}i}{-36}$$

Divide -12+24i\sqrt{5} by -36.

$$x=\frac{-2\sqrt{5}i+1}{3}$$

Now solve the equation x=\frac{-12±24\sqrt{5}i}{-36} when ± is minus. Subtract 24i\sqrt{5} from -12.

$$x=\frac{-24\sqrt{5}i-12}{-36}$$

Divide -12-24i\sqrt{5} by -36.

$$x=\frac{1+2\sqrt{5}i}{3}$$

The equation is now solved.

$$x=\frac{-2\sqrt{5}i+1}{3} x=\frac{1+2\sqrt{5}i}{3}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-1\right)^{2}.

$$7-2\left(9x^{2}-6x+1\right)=47$$

Use the distributive property to multiply -2 by 9x^{2}-6x+1.

$$7-18x^{2}+12x-2=47$$

Subtract 2 from 7 to get 5.

$$5-18x^{2}+12x=47$$

Subtract 5 from both sides.

$$-18x^{2}+12x=47-5$$

Subtract 5 from 47 to get 42.

$$-18x^{2}+12x=42$$

Divide both sides by -18.

$$\frac{-18x^{2}+12x}{-18}=\frac{42}{-18}$$

Dividing by -18 undoes the multiplication by -18.

$$x^{2}+\frac{12}{-18}x=\frac{42}{-18}$$

Reduce the fraction \frac{12}{-18} to lowest terms by extracting and canceling out 6.

$$x^{2}-\frac{2}{3}x=\frac{42}{-18}$$

Reduce the fraction \frac{42}{-18} to lowest terms by extracting and canceling out 6.

$$x^{2}-\frac{2}{3}x=-\frac{7}{3}$$

Divide -\frac{2}{3}, the coefficient of the x term, by 2 to get -\frac{1}{3}. Then add the square of -\frac{1}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{2}{3}x+\left(-\frac{1}{3}\right)^{2}=-\frac{7}{3}+\left(-\frac{1}{3}\right)^{2}$$

Square -\frac{1}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{2}{3}x+\frac{1}{9}=-\frac{7}{3}+\frac{1}{9}$$

Add -\frac{7}{3} to \frac{1}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{2}{3}x+\frac{1}{9}=-\frac{20}{9}$$

Factor x^{2}-\frac{2}{3}x+\frac{1}{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}.

$$\left(x-\frac{1}{3}\right)^{2}=-\frac{20}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{3}\right)^{2}}=\sqrt{-\frac{20}{9}}$$

Simplify.

$$x-\frac{1}{3}=\frac{2\sqrt{5}i}{3} x-\frac{1}{3}=-\frac{2\sqrt{5}i}{3}$$

Add \frac{1}{3} to both sides of the equation.

$$x=\frac{1+2\sqrt{5}i}{3} x=\frac{-2\sqrt{5}i+1}{3}$$