Solve =4x+6quad(9)(x-1)^2+2(x-1)-3=0

$$= 4 x + 6 \quad ( 9 ) ( x - 1 ) ^ { 2 } + 2 ( x - 1 ) - 3 = 0$$

Solve for x (complex solution)

$$x=\frac{17+\sqrt{5}i}{18}\approx 0.944444444+0.124225999i$$

$$x=\frac{-\sqrt{5}i+17}{18}\approx 0.944444444-0.124225999i$$

Steps Using the Quadratic Formula

Multiply 6 and 9 to get 54.

$$4x+54\left(x-1\right)^{2}+2\left(x-1\right)-3=0$$

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

$$4x+54\left(x^{2}-2x+1\right)+2\left(x-1\right)-3=0$$

Use the distributive property to multiply 54 by x^{2}-2x+1.

$$4x+54x^{2}-108x+54+2\left(x-1\right)-3=0$$

Combine 4x and -108x to get -104x.

$$-104x+54x^{2}+54+2\left(x-1\right)-3=0$$

Use the distributive property to multiply 2 by x-1.

$$-104x+54x^{2}+54+2x-2-3=0$$

Combine -104x and 2x to get -102x.

$$-102x+54x^{2}+54-2-3=0$$

Subtract 2 from 54 to get 52.

$$-102x+54x^{2}+52-3=0$$

Subtract 3 from 52 to get 49.

$$-102x+54x^{2}+49=0$$

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.

$$54x^{2}-102x+49=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 54 for a, -102 for b, and 49 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-102\right)±\sqrt{\left(-102\right)^{2}-4\times 54\times 49}}{2\times 54}$$

Square -102.

$$x=\frac{-\left(-102\right)±\sqrt{10404-4\times 54\times 49}}{2\times 54}$$

Multiply -4 times 54.

$$x=\frac{-\left(-102\right)±\sqrt{10404-216\times 49}}{2\times 54}$$

Multiply -216 times 49.

$$x=\frac{-\left(-102\right)±\sqrt{10404-10584}}{2\times 54}$$

Add 10404 to -10584.

$$x=\frac{-\left(-102\right)±\sqrt{-180}}{2\times 54}$$

Take the square root of -180.

$$x=\frac{-\left(-102\right)±6\sqrt{5}i}{2\times 54}$$

The opposite of -102 is 102.

$$x=\frac{102±6\sqrt{5}i}{2\times 54}$$

Multiply 2 times 54.

$$x=\frac{102±6\sqrt{5}i}{108}$$

Now solve the equation x=\frac{102±6\sqrt{5}i}{108} when ± is plus. Add 102 to 6i\sqrt{5}.

$$x=\frac{102+6\sqrt{5}i}{108}$$

Divide 102+6i\sqrt{5} by 108.

$$x=\frac{17+\sqrt{5}i}{18}$$

Now solve the equation x=\frac{102±6\sqrt{5}i}{108} when ± is minus. Subtract 6i\sqrt{5} from 102.

$$x=\frac{-6\sqrt{5}i+102}{108}$$

Divide 102-6i\sqrt{5} by 108.

$$x=\frac{-\sqrt{5}i+17}{18}$$

The equation is now solved.

$$x=\frac{17+\sqrt{5}i}{18} x=\frac{-\sqrt{5}i+17}{18}$$

Steps for Completing the Square

Multiply 6 and 9 to get 54.

$$4x+54\left(x-1\right)^{2}+2\left(x-1\right)-3=0$$

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

$$4x+54\left(x^{2}-2x+1\right)+2\left(x-1\right)-3=0$$

Use the distributive property to multiply 54 by x^{2}-2x+1.

$$4x+54x^{2}-108x+54+2\left(x-1\right)-3=0$$

Combine 4x and -108x to get -104x.

$$-104x+54x^{2}+54+2\left(x-1\right)-3=0$$

Use the distributive property to multiply 2 by x-1.

$$-104x+54x^{2}+54+2x-2-3=0$$

Combine -104x and 2x to get -102x.

$$-102x+54x^{2}+54-2-3=0$$

Subtract 2 from 54 to get 52.

$$-102x+54x^{2}+52-3=0$$

Subtract 3 from 52 to get 49.

$$-102x+54x^{2}+49=0$$

Subtract 49 from both sides. Anything subtracted from zero gives its negation.

$$-102x+54x^{2}=-49$$

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.

$$54x^{2}-102x=-49$$

Divide both sides by 54.

$$\frac{54x^{2}-102x}{54}=-\frac{49}{54}$$

Dividing by 54 undoes the multiplication by 54.

$$x^{2}+\left(-\frac{102}{54}\right)x=-\frac{49}{54}$$

Reduce the fraction \frac{-102}{54} to lowest terms by extracting and canceling out 6.

$$x^{2}-\frac{17}{9}x=-\frac{49}{54}$$

Divide -\frac{17}{9}, the coefficient of the x term, by 2 to get -\frac{17}{18}. Then add the square of -\frac{17}{18} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{17}{9}x+\left(-\frac{17}{18}\right)^{2}=-\frac{49}{54}+\left(-\frac{17}{18}\right)^{2}$$

Square -\frac{17}{18} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{17}{9}x+\frac{289}{324}=-\frac{49}{54}+\frac{289}{324}$$

Add -\frac{49}{54} to \frac{289}{324} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{17}{9}x+\frac{289}{324}=-\frac{5}{324}$$

Factor x^{2}-\frac{17}{9}x+\frac{289}{324}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{17}{18}\right)^{2}=-\frac{5}{324}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{17}{18}\right)^{2}}=\sqrt{-\frac{5}{324}}$$

Simplify.

$$x-\frac{17}{18}=\frac{\sqrt{5}i}{18} x-\frac{17}{18}=-\frac{\sqrt{5}i}{18}$$

Add \frac{17}{18} to both sides of the equation.

$$x=\frac{17+\sqrt{5}i}{18} x=\frac{-\sqrt{5}i+17}{18}$$