Solve -4/3x+4=(x-1)^2-4

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

Solve for x

$$x = -\frac{7}{3} = -2\frac{1}{3} \approx -2.333333333$$

$$x=3$$

Steps Using the Quadratic Formula

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

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

Subtract 4 from 1 to get -3.

$$-\frac{4}{3}x+4=x^{2}-2x-3$$

Subtract x^{2} from both sides.

$$-\frac{4}{3}x+4-x^{2}=-2x-3$$

Add 2x to both sides.

$$-\frac{4}{3}x+4-x^{2}+2x=-3$$

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

$$\frac{2}{3}x+4-x^{2}=-3$$

Add 3 to both sides.

$$\frac{2}{3}x+4-x^{2}+3=0$$

Add 4 and 3 to get 7.

$$\frac{2}{3}x+7-x^{2}=0$$

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.

$$-x^{2}+\frac{2}{3}x+7=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, \frac{2}{3} for b, and 7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\frac{2}{3}±\sqrt{\left(\frac{2}{3}\right)^{2}-4\left(-1\right)\times 7}}{2\left(-1\right)}$$

Square \frac{2}{3} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\frac{2}{3}±\sqrt{\frac{4}{9}-4\left(-1\right)\times 7}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\frac{2}{3}±\sqrt{\frac{4}{9}+4\times 7}}{2\left(-1\right)}$$

Multiply 4 times 7.

$$x=\frac{-\frac{2}{3}±\sqrt{\frac{4}{9}+28}}{2\left(-1\right)}$$

Add \frac{4}{9} to 28.

$$x=\frac{-\frac{2}{3}±\sqrt{\frac{256}{9}}}{2\left(-1\right)}$$

Take the square root of \frac{256}{9}.

$$x=\frac{-\frac{2}{3}±\frac{16}{3}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-\frac{2}{3}±\frac{16}{3}}{-2}$$

Now solve the equation x=\frac{-\frac{2}{3}±\frac{16}{3}}{-2} when ± is plus. Add -\frac{2}{3} to \frac{16}{3} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{\frac{14}{3}}{-2}$$

Divide \frac{14}{3} by -2.

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

Now solve the equation x=\frac{-\frac{2}{3}±\frac{16}{3}}{-2} when ± is minus. Subtract \frac{16}{3} from -\frac{2}{3} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

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

Divide -6 by -2.

$$x=3$$

The equation is now solved.

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

Steps for Completing the Square

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

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

Subtract 4 from 1 to get -3.

$$-\frac{4}{3}x+4=x^{2}-2x-3$$

Subtract x^{2} from both sides.

$$-\frac{4}{3}x+4-x^{2}=-2x-3$$

Add 2x to both sides.

$$-\frac{4}{3}x+4-x^{2}+2x=-3$$

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

$$\frac{2}{3}x+4-x^{2}=-3$$

Subtract 4 from both sides.

$$\frac{2}{3}x-x^{2}=-3-4$$

Subtract 4 from -3 to get -7.

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

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.

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

Divide both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

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

Divide \frac{2}{3} by -1.

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

Divide -7 by -1.

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

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.

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

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

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

Add 7 to \frac{1}{9}.

$$x^{2}-\frac{2}{3}x+\frac{1}{9}=\frac{64}{9}$$

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}.

$$\left(x-\frac{1}{3}\right)^{2}=\frac{64}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{3}\right)^{2}}=\sqrt{\frac{64}{9}}$$

Simplify.

$$x-\frac{1}{3}=\frac{8}{3} x-\frac{1}{3}=-\frac{8}{3}$$

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

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