Solve 3(x-5)^2-4(x-5)+2=0

$$3 ( x - 5 ) ^ { 2 } - 4 ( x - 5 ) + 2 = 0$$

Solve for x (complex solution)

$$x=\frac{17+\sqrt{2}i}{3}\approx 5.666666667+0.471404521i$$

$$x=\frac{-\sqrt{2}i+17}{3}\approx 5.666666667-0.471404521i$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-5\right)^{2}.

$$3\left(x^{2}-10x+25\right)-4\left(x-5\right)+2=0$$

Use the distributive property to multiply 3 by x^{2}-10x+25.

$$3x^{2}-30x+75-4\left(x-5\right)+2=0$$

Use the distributive property to multiply -4 by x-5.

$$3x^{2}-30x+75-4x+20+2=0$$

Combine -30x and -4x to get -34x.

$$3x^{2}-34x+75+20+2=0$$

Add 75 and 20 to get 95.

$$3x^{2}-34x+95+2=0$$

Add 95 and 2 to get 97.

$$3x^{2}-34x+97=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -34 for b, and 97 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-34\right)±\sqrt{\left(-34\right)^{2}-4\times 3\times 97}}{2\times 3}$$

Square -34.

$$x=\frac{-\left(-34\right)±\sqrt{1156-4\times 3\times 97}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-34\right)±\sqrt{1156-12\times 97}}{2\times 3}$$

Multiply -12 times 97.

$$x=\frac{-\left(-34\right)±\sqrt{1156-1164}}{2\times 3}$$

Add 1156 to -1164.

$$x=\frac{-\left(-34\right)±\sqrt{-8}}{2\times 3}$$

Take the square root of -8.

$$x=\frac{-\left(-34\right)±2\sqrt{2}i}{2\times 3}$$

The opposite of -34 is 34.

$$x=\frac{34±2\sqrt{2}i}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{34±2\sqrt{2}i}{6}$$

Now solve the equation x=\frac{34±2\sqrt{2}i}{6} when ± is plus. Add 34 to 2i\sqrt{2}.

$$x=\frac{34+2\sqrt{2}i}{6}$$

Divide 34+2i\sqrt{2} by 6.

$$x=\frac{17+\sqrt{2}i}{3}$$

Now solve the equation x=\frac{34±2\sqrt{2}i}{6} when ± is minus. Subtract 2i\sqrt{2} from 34.

$$x=\frac{-2\sqrt{2}i+34}{6}$$

Divide 34-2i\sqrt{2} by 6.

$$x=\frac{-\sqrt{2}i+17}{3}$$

The equation is now solved.

$$x=\frac{17+\sqrt{2}i}{3} x=\frac{-\sqrt{2}i+17}{3}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-5\right)^{2}.

$$3\left(x^{2}-10x+25\right)-4\left(x-5\right)+2=0$$

Use the distributive property to multiply 3 by x^{2}-10x+25.

$$3x^{2}-30x+75-4\left(x-5\right)+2=0$$

Use the distributive property to multiply -4 by x-5.

$$3x^{2}-30x+75-4x+20+2=0$$

Combine -30x and -4x to get -34x.

$$3x^{2}-34x+75+20+2=0$$

Add 75 and 20 to get 95.

$$3x^{2}-34x+95+2=0$$

Add 95 and 2 to get 97.

$$3x^{2}-34x+97=0$$

Subtract 97 from both sides. Anything subtracted from zero gives its negation.

$$3x^{2}-34x=-97$$

Divide both sides by 3.

$$\frac{3x^{2}-34x}{3}=-\frac{97}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}-\frac{34}{3}x=-\frac{97}{3}$$

Divide -\frac{34}{3}, the coefficient of the x term, by 2 to get -\frac{17}{3}. Then add the square of -\frac{17}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{34}{3}x+\left(-\frac{17}{3}\right)^{2}=-\frac{97}{3}+\left(-\frac{17}{3}\right)^{2}$$

Square -\frac{17}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{34}{3}x+\frac{289}{9}=-\frac{97}{3}+\frac{289}{9}$$

Add -\frac{97}{3} to \frac{289}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{34}{3}x+\frac{289}{9}=-\frac{2}{9}$$

Factor x^{2}-\frac{34}{3}x+\frac{289}{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}.

$$\left(x-\frac{17}{3}\right)^{2}=-\frac{2}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{17}{3}\right)^{2}}=\sqrt{-\frac{2}{9}}$$

Simplify.

$$x-\frac{17}{3}=\frac{\sqrt{2}i}{3} x-\frac{17}{3}=-\frac{\sqrt{2}i}{3}$$

Add \frac{17}{3} to both sides of the equation.

$$x=\frac{17+\sqrt{2}i}{3} x=\frac{-\sqrt{2}i+17}{3}$$