Solve 1/3*x*(x-2)=5

$$\frac { 1 } { 3 } \cdot x \cdot ( x - 2 ) = 5$$

Solve for x

$$x=-3$$

$$x=5$$

Steps Using the Quadratic Formula

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

$$\frac{1}{3}xx+\frac{1}{3}x\left(-2\right)=5$$

Multiply x and x to get x^{2}.

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

Multiply \frac{1}{3} and -2 to get \frac{-2}{3}.

$$\frac{1}{3}x^{2}+\frac{-2}{3}x=5$$

Fraction \frac{-2}{3} can be rewritten as -\frac{2}{3} by extracting the negative sign.

$$\frac{1}{3}x^{2}-\frac{2}{3}x=5$$

Subtract 5 from both sides.

$$\frac{1}{3}x^{2}-\frac{2}{3}x-5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{1}{3} for a, -\frac{2}{3} for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-\frac{2}{3}\right)±\sqrt{\left(-\frac{2}{3}\right)^{2}-4\times \frac{1}{3}\left(-5\right)}}{2\times \frac{1}{3}}$$

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

$$x=\frac{-\left(-\frac{2}{3}\right)±\sqrt{\frac{4}{9}-4\times \frac{1}{3}\left(-5\right)}}{2\times \frac{1}{3}}$$

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

$$x=\frac{-\left(-\frac{2}{3}\right)±\sqrt{\frac{4}{9}-\frac{4}{3}\left(-5\right)}}{2\times \frac{1}{3}}$$

Multiply -\frac{4}{3} times -5.

$$x=\frac{-\left(-\frac{2}{3}\right)±\sqrt{\frac{4}{9}+\frac{20}{3}}}{2\times \frac{1}{3}}$$

Add \frac{4}{9} to \frac{20}{3} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

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

Take the square root of \frac{64}{9}.

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

The opposite of -\frac{2}{3} is \frac{2}{3}.

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

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

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

Now solve the equation x=\frac{\frac{2}{3}±\frac{8}{3}}{\frac{2}{3}} when ± is plus. Add \frac{2}{3} to \frac{8}{3} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{\frac{10}{3}}{\frac{2}{3}}$$

Divide \frac{10}{3} by \frac{2}{3} by multiplying \frac{10}{3} by the reciprocal of \frac{2}{3}.

$$x=5$$

Now solve the equation x=\frac{\frac{2}{3}±\frac{8}{3}}{\frac{2}{3}} when ± is minus. Subtract \frac{8}{3} from \frac{2}{3} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

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

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

$$x=-3$$

The equation is now solved.

$$x=5 x=-3$$

Steps for Completing the Square

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

$$\frac{1}{3}xx+\frac{1}{3}x\left(-2\right)=5$$

Multiply x and x to get x^{2}.

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

Multiply \frac{1}{3} and -2 to get \frac{-2}{3}.

$$\frac{1}{3}x^{2}+\frac{-2}{3}x=5$$

Fraction \frac{-2}{3} can be rewritten as -\frac{2}{3} by extracting the negative sign.

$$\frac{1}{3}x^{2}-\frac{2}{3}x=5$$

Multiply both sides by 3.

$$\frac{\frac{1}{3}x^{2}-\frac{2}{3}x}{\frac{1}{3}}=\frac{5}{\frac{1}{3}}$$

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

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

Divide -\frac{2}{3} by \frac{1}{3} by multiplying -\frac{2}{3} by the reciprocal of \frac{1}{3}.

$$x^{2}-2x=\frac{5}{\frac{1}{3}}$$

Divide 5 by \frac{1}{3} by multiplying 5 by the reciprocal of \frac{1}{3}.

$$x^{2}-2x=15$$

Divide -2, the coefficient of the x term, by 2 to get -1. Then add the square of -1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-2x+1=15+1$$

Add 15 to 1.

$$x^{2}-2x+1=16$$

Factor x^{2}-2x+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-1\right)^{2}=16$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-1\right)^{2}}=\sqrt{16}$$

Simplify.

$$x-1=4 x-1=-4$$

Add 1 to both sides of the equation.

$$x=5 x=-3$$