Solve 0=1/3left(x-4right)^2-3

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

Solve for x

$$x=7$$

$$x=1$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-4\right)^{2}.

$$0=\frac{1}{3}\left(x^{2}-8x+16\right)-3$$

Use the distributive property to multiply \frac{1}{3} by x^{2}-8x+16.

$$0=\frac{1}{3}x^{2}-\frac{8}{3}x+\frac{16}{3}-3$$

Subtract 3 from \frac{16}{3} to get \frac{7}{3}.

$$0=\frac{1}{3}x^{2}-\frac{8}{3}x+\frac{7}{3}$$

Swap sides so that all variable terms are on the left hand side.

$$\frac{1}{3}x^{2}-\frac{8}{3}x+\frac{7}{3}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{1}{3} for a, -\frac{8}{3} for b, and \frac{7}{3} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-\frac{8}{3}\right)±\sqrt{\left(-\frac{8}{3}\right)^{2}-4\times \frac{1}{3}\times \frac{7}{3}}}{2\times \frac{1}{3}}$$

Square -\frac{8}{3} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\left(-\frac{8}{3}\right)±\sqrt{\frac{64}{9}-4\times \frac{1}{3}\times \frac{7}{3}}}{2\times \frac{1}{3}}$$

Multiply -4 times \frac{1}{3}.

$$x=\frac{-\left(-\frac{8}{3}\right)±\sqrt{\frac{64}{9}-\frac{4}{3}\times \frac{7}{3}}}{2\times \frac{1}{3}}$$

Multiply -\frac{4}{3} times \frac{7}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\left(-\frac{8}{3}\right)±\sqrt{\frac{64-28}{9}}}{2\times \frac{1}{3}}$$

Add \frac{64}{9} to -\frac{28}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\left(-\frac{8}{3}\right)±\sqrt{4}}{2\times \frac{1}{3}}$$

Take the square root of 4.

$$x=\frac{-\left(-\frac{8}{3}\right)±2}{2\times \frac{1}{3}}$$

The opposite of -\frac{8}{3} is \frac{8}{3}.

$$x=\frac{\frac{8}{3}±2}{2\times \frac{1}{3}}$$

Multiply 2 times \frac{1}{3}.

$$x=\frac{\frac{8}{3}±2}{\frac{2}{3}}$$

Now solve the equation x=\frac{\frac{8}{3}±2}{\frac{2}{3}} when ± is plus. Add \frac{8}{3} to 2.

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

Divide \frac{14}{3} by \frac{2}{3} by multiplying \frac{14}{3} by the reciprocal of \frac{2}{3}.

$$x=7$$

Now solve the equation x=\frac{\frac{8}{3}±2}{\frac{2}{3}} when ± is minus. Subtract 2 from \frac{8}{3}.

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

Divide \frac{2}{3} by \frac{2}{3} by multiplying \frac{2}{3} by the reciprocal of \frac{2}{3}.

$$x=1$$

The equation is now solved.

$$x=7 x=1$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-4\right)^{2}.

$$0=\frac{1}{3}\left(x^{2}-8x+16\right)-3$$

Use the distributive property to multiply \frac{1}{3} by x^{2}-8x+16.

$$0=\frac{1}{3}x^{2}-\frac{8}{3}x+\frac{16}{3}-3$$

Subtract 3 from \frac{16}{3} to get \frac{7}{3}.

$$0=\frac{1}{3}x^{2}-\frac{8}{3}x+\frac{7}{3}$$

Swap sides so that all variable terms are on the left hand side.

$$\frac{1}{3}x^{2}-\frac{8}{3}x+\frac{7}{3}=0$$

Subtract \frac{7}{3} from both sides. Anything subtracted from zero gives its negation.

$$\frac{1}{3}x^{2}-\frac{8}{3}x=-\frac{7}{3}$$

Multiply both sides by 3.

$$\frac{\frac{1}{3}x^{2}-\frac{8}{3}x}{\frac{1}{3}}=-\frac{\frac{7}{3}}{\frac{1}{3}}$$

Dividing by \frac{1}{3} undoes the multiplication by \frac{1}{3}.

$$x^{2}+\left(-\frac{\frac{8}{3}}{\frac{1}{3}}\right)x=-\frac{\frac{7}{3}}{\frac{1}{3}}$$

Divide -\frac{8}{3} by \frac{1}{3} by multiplying -\frac{8}{3} by the reciprocal of \frac{1}{3}.

$$x^{2}-8x=-\frac{\frac{7}{3}}{\frac{1}{3}}$$

Divide -\frac{7}{3} by \frac{1}{3} by multiplying -\frac{7}{3} by the reciprocal of \frac{1}{3}.

$$x^{2}-8x=-7$$

Divide -8, the coefficient of the x term, by 2 to get -4. Then add the square of -4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-8x+\left(-4\right)^{2}=-7+\left(-4\right)^{2}$$

Square -4.

$$x^{2}-8x+16=-7+16$$

Add -7 to 16.

$$x^{2}-8x+16=9$$

Factor x^{2}-8x+16. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-4\right)^{2}=9$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-4\right)^{2}}=\sqrt{9}$$

Simplify.

$$x-4=3 x-4=-3$$

Add 4 to both sides of the equation.

$$x=7 x=1$$