Solve (2x-1/2)^2+4/3(x-1/6)+2=(3x-1/2)(3x+frac{1}{2})+12x

$$( 2 x - \frac { 1 } { 2 } ) ^ { 2 } + \frac { 4 } { 3 } ( x - \frac { 1 } { 6 } ) + 2 = ( 3 x - \frac { 1 } { 2 } ) ( 3 x + \frac { 1 } { 2 } ) + 12 x$$

Solve for x

$$x=\frac{\sqrt{206}}{10}-\frac{19}{15}\approx 0.168603343$$

$$x=-\frac{\sqrt{206}}{10}-\frac{19}{15}\approx -2.701936676$$

Steps Using the Quadratic Formula

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

$$4x^{2}-2x+\frac{1}{4}+\frac{4}{3}\left(x-\frac{1}{6}\right)+2=\left(3x-\frac{1}{2}\right)\left(3x+\frac{1}{2}\right)+12x$$

Use the distributive property to multiply \frac{4}{3} by x-\frac{1}{6}.

$$4x^{2}-2x+\frac{1}{4}+\frac{4}{3}x-\frac{2}{9}+2=\left(3x-\frac{1}{2}\right)\left(3x+\frac{1}{2}\right)+12x$$

Combine -2x and \frac{4}{3}x to get -\frac{2}{3}x.

$$4x^{2}-\frac{2}{3}x+\frac{1}{4}-\frac{2}{9}+2=\left(3x-\frac{1}{2}\right)\left(3x+\frac{1}{2}\right)+12x$$

Subtract \frac{2}{9} from \frac{1}{4} to get \frac{1}{36}.

$$4x^{2}-\frac{2}{3}x+\frac{1}{36}+2=\left(3x-\frac{1}{2}\right)\left(3x+\frac{1}{2}\right)+12x$$

Add \frac{1}{36} and 2 to get \frac{73}{36}.

$$4x^{2}-\frac{2}{3}x+\frac{73}{36}=\left(3x-\frac{1}{2}\right)\left(3x+\frac{1}{2}\right)+12x$$

Consider \left(3x-\frac{1}{2}\right)\left(3x+\frac{1}{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 \frac{1}{2}.

$$4x^{2}-\frac{2}{3}x+\frac{73}{36}=\left(3x\right)^{2}-\frac{1}{4}+12x$$

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

$$4x^{2}-\frac{2}{3}x+\frac{73}{36}=3^{2}x^{2}-\frac{1}{4}+12x$$

Calculate 3 to the power of 2 and get 9.

$$4x^{2}-\frac{2}{3}x+\frac{73}{36}=9x^{2}-\frac{1}{4}+12x$$

Subtract 9x^{2} from both sides.

$$4x^{2}-\frac{2}{3}x+\frac{73}{36}-9x^{2}=-\frac{1}{4}+12x$$

Combine 4x^{2} and -9x^{2} to get -5x^{2}.

$$-5x^{2}-\frac{2}{3}x+\frac{73}{36}=-\frac{1}{4}+12x$$

Subtract -\frac{1}{4} from both sides.

$$-5x^{2}-\frac{2}{3}x+\frac{73}{36}-\left(-\frac{1}{4}\right)=12x$$

The opposite of -\frac{1}{4} is \frac{1}{4}.

$$-5x^{2}-\frac{2}{3}x+\frac{73}{36}+\frac{1}{4}=12x$$

Subtract 12x from both sides.

$$-5x^{2}-\frac{2}{3}x+\frac{73}{36}+\frac{1}{4}-12x=0$$

Add \frac{73}{36} and \frac{1}{4} to get \frac{41}{18}.

$$-5x^{2}-\frac{2}{3}x+\frac{41}{18}-12x=0$$

Combine -\frac{2}{3}x and -12x to get -\frac{38}{3}x.

$$-5x^{2}-\frac{38}{3}x+\frac{41}{18}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -5 for a, -\frac{38}{3} for b, and \frac{41}{18} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-\frac{38}{3}\right)±\sqrt{\left(-\frac{38}{3}\right)^{2}-4\left(-5\right)\times \frac{41}{18}}}{2\left(-5\right)}$$

Square -\frac{38}{3} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\left(-\frac{38}{3}\right)±\sqrt{\frac{1444}{9}-4\left(-5\right)\times \frac{41}{18}}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$x=\frac{-\left(-\frac{38}{3}\right)±\sqrt{\frac{1444}{9}+20\times \frac{41}{18}}}{2\left(-5\right)}$$

Multiply 20 times \frac{41}{18}.

$$x=\frac{-\left(-\frac{38}{3}\right)±\sqrt{\frac{1444+410}{9}}}{2\left(-5\right)}$$

Add \frac{1444}{9} to \frac{410}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\left(-\frac{38}{3}\right)±\sqrt{206}}{2\left(-5\right)}$$

The opposite of -\frac{38}{3} is \frac{38}{3}.

$$x=\frac{\frac{38}{3}±\sqrt{206}}{2\left(-5\right)}$$

Multiply 2 times -5.

$$x=\frac{\frac{38}{3}±\sqrt{206}}{-10}$$

Now solve the equation x=\frac{\frac{38}{3}±\sqrt{206}}{-10} when ± is plus. Add \frac{38}{3} to \sqrt{206}.

$$x=\frac{\sqrt{206}+\frac{38}{3}}{-10}$$

Divide \frac{38}{3}+\sqrt{206} by -10.

$$x=-\frac{\sqrt{206}}{10}-\frac{19}{15}$$

Now solve the equation x=\frac{\frac{38}{3}±\sqrt{206}}{-10} when ± is minus. Subtract \sqrt{206} from \frac{38}{3}.

$$x=\frac{\frac{38}{3}-\sqrt{206}}{-10}$$

Divide \frac{38}{3}-\sqrt{206} by -10.

$$x=\frac{\sqrt{206}}{10}-\frac{19}{15}$$

The equation is now solved.

$$x=-\frac{\sqrt{206}}{10}-\frac{19}{15} x=\frac{\sqrt{206}}{10}-\frac{19}{15}$$

Steps for Completing the Square

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

$$4x^{2}-2x+\frac{1}{4}+\frac{4}{3}\left(x-\frac{1}{6}\right)+2=\left(3x-\frac{1}{2}\right)\left(3x+\frac{1}{2}\right)+12x$$

Use the distributive property to multiply \frac{4}{3} by x-\frac{1}{6}.

$$4x^{2}-2x+\frac{1}{4}+\frac{4}{3}x-\frac{2}{9}+2=\left(3x-\frac{1}{2}\right)\left(3x+\frac{1}{2}\right)+12x$$

Combine -2x and \frac{4}{3}x to get -\frac{2}{3}x.

$$4x^{2}-\frac{2}{3}x+\frac{1}{4}-\frac{2}{9}+2=\left(3x-\frac{1}{2}\right)\left(3x+\frac{1}{2}\right)+12x$$

Subtract \frac{2}{9} from \frac{1}{4} to get \frac{1}{36}.

$$4x^{2}-\frac{2}{3}x+\frac{1}{36}+2=\left(3x-\frac{1}{2}\right)\left(3x+\frac{1}{2}\right)+12x$$

Add \frac{1}{36} and 2 to get \frac{73}{36}.

$$4x^{2}-\frac{2}{3}x+\frac{73}{36}=\left(3x-\frac{1}{2}\right)\left(3x+\frac{1}{2}\right)+12x$$

Consider \left(3x-\frac{1}{2}\right)\left(3x+\frac{1}{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 \frac{1}{2}.

$$4x^{2}-\frac{2}{3}x+\frac{73}{36}=\left(3x\right)^{2}-\frac{1}{4}+12x$$

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

$$4x^{2}-\frac{2}{3}x+\frac{73}{36}=3^{2}x^{2}-\frac{1}{4}+12x$$

Calculate 3 to the power of 2 and get 9.

$$4x^{2}-\frac{2}{3}x+\frac{73}{36}=9x^{2}-\frac{1}{4}+12x$$

Subtract 9x^{2} from both sides.

$$4x^{2}-\frac{2}{3}x+\frac{73}{36}-9x^{2}=-\frac{1}{4}+12x$$

Combine 4x^{2} and -9x^{2} to get -5x^{2}.

$$-5x^{2}-\frac{2}{3}x+\frac{73}{36}=-\frac{1}{4}+12x$$

Subtract 12x from both sides.

$$-5x^{2}-\frac{2}{3}x+\frac{73}{36}-12x=-\frac{1}{4}$$

Combine -\frac{2}{3}x and -12x to get -\frac{38}{3}x.

$$-5x^{2}-\frac{38}{3}x+\frac{73}{36}=-\frac{1}{4}$$

Subtract \frac{73}{36} from both sides.

$$-5x^{2}-\frac{38}{3}x=-\frac{1}{4}-\frac{73}{36}$$

Subtract \frac{73}{36} from -\frac{1}{4} to get -\frac{41}{18}.

$$-5x^{2}-\frac{38}{3}x=-\frac{41}{18}$$

Divide both sides by -5.

$$\frac{-5x^{2}-\frac{38}{3}x}{-5}=-\frac{\frac{41}{18}}{-5}$$

Dividing by -5 undoes the multiplication by -5.

$$x^{2}+\left(-\frac{\frac{38}{3}}{-5}\right)x=-\frac{\frac{41}{18}}{-5}$$

Divide -\frac{38}{3} by -5.

$$x^{2}+\frac{38}{15}x=-\frac{\frac{41}{18}}{-5}$$

Divide -\frac{41}{18} by -5.

$$x^{2}+\frac{38}{15}x=\frac{41}{90}$$

Divide \frac{38}{15}, the coefficient of the x term, by 2 to get \frac{19}{15}. Then add the square of \frac{19}{15} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{38}{15}x+\left(\frac{19}{15}\right)^{2}=\frac{41}{90}+\left(\frac{19}{15}\right)^{2}$$

Square \frac{19}{15} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{38}{15}x+\frac{361}{225}=\frac{41}{90}+\frac{361}{225}$$

Add \frac{41}{90} to \frac{361}{225} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{38}{15}x+\frac{361}{225}=\frac{103}{50}$$

Factor x^{2}+\frac{38}{15}x+\frac{361}{225}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{19}{15}\right)^{2}=\frac{103}{50}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{19}{15}\right)^{2}}=\sqrt{\frac{103}{50}}$$

Simplify.

$$x+\frac{19}{15}=\frac{\sqrt{206}}{10} x+\frac{19}{15}=-\frac{\sqrt{206}}{10}$$

Subtract \frac{19}{15} from both sides of the equation.

$$x=\frac{\sqrt{206}}{10}-\frac{19}{15} x=-\frac{\sqrt{206}}{10}-\frac{19}{15}$$