Solve 0=x*(-6/5x+4)

$$0 = x \cdot ( - \frac { 6 } { 5 } x + 4 )$$

Solve for x

$$x = \frac{10}{3} = 3\frac{1}{3} \approx 3.333333333$$

$$x=0$$

Steps Using Factoring

Use the distributive property to multiply x by -\frac{6}{5}x+4.

$$0=x\left(-\frac{6}{5}\right)x+4x$$

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

$$0=x^{2}\left(-\frac{6}{5}\right)+4x$$

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

$$x^{2}\left(-\frac{6}{5}\right)+4x=0$$

Factor out x.

$$x\left(-\frac{6}{5}x+4\right)=0$$

To find equation solutions, solve x=0 and -\frac{6x}{5}+4=0.

$$x=0 x=\frac{10}{3}$$

Steps Using the Quadratic Formula

Use the distributive property to multiply x by -\frac{6}{5}x+4.

$$0=x\left(-\frac{6}{5}\right)x+4x$$

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

$$0=x^{2}\left(-\frac{6}{5}\right)+4x$$

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

$$x^{2}\left(-\frac{6}{5}\right)+4x=0$$

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.

$$-\frac{6}{5}x^{2}+4x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{6}{5} for a, 4 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-4±\sqrt{4^{2}}}{2\left(-\frac{6}{5}\right)}$$

Take the square root of 4^{2}.

$$x=\frac{-4±4}{2\left(-\frac{6}{5}\right)}$$

Multiply 2 times -\frac{6}{5}.

$$x=\frac{-4±4}{-\frac{12}{5}}$$

Now solve the equation x=\frac{-4±4}{-\frac{12}{5}} when ± is plus. Add -4 to 4.

$$x=\frac{0}{-\frac{12}{5}}$$

Divide 0 by -\frac{12}{5} by multiplying 0 by the reciprocal of -\frac{12}{5}.

$$x=0$$

Now solve the equation x=\frac{-4±4}{-\frac{12}{5}} when ± is minus. Subtract 4 from -4.

$$x=-\frac{8}{-\frac{12}{5}}$$

Divide -8 by -\frac{12}{5} by multiplying -8 by the reciprocal of -\frac{12}{5}.

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

The equation is now solved.

$$x=0 x=\frac{10}{3}$$

Steps for Completing the Square

Use the distributive property to multiply x by -\frac{6}{5}x+4.

$$0=x\left(-\frac{6}{5}\right)x+4x$$

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

$$0=x^{2}\left(-\frac{6}{5}\right)+4x$$

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

$$x^{2}\left(-\frac{6}{5}\right)+4x=0$$

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.

$$-\frac{6}{5}x^{2}+4x=0$$

Divide both sides of the equation by -\frac{6}{5}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{-\frac{6}{5}x^{2}+4x}{-\frac{6}{5}}=\frac{0}{-\frac{6}{5}}$$

Dividing by -\frac{6}{5} undoes the multiplication by -\frac{6}{5}.

$$x^{2}+\frac{4}{-\frac{6}{5}}x=\frac{0}{-\frac{6}{5}}$$

Divide 4 by -\frac{6}{5} by multiplying 4 by the reciprocal of -\frac{6}{5}.

$$x^{2}-\frac{10}{3}x=\frac{0}{-\frac{6}{5}}$$

Divide 0 by -\frac{6}{5} by multiplying 0 by the reciprocal of -\frac{6}{5}.

$$x^{2}-\frac{10}{3}x=0$$

Divide -\frac{10}{3}, the coefficient of the x term, by 2 to get -\frac{5}{3}. Then add the square of -\frac{5}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{10}{3}x+\left(-\frac{5}{3}\right)^{2}=\left(-\frac{5}{3}\right)^{2}$$

Square -\frac{5}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{10}{3}x+\frac{25}{9}=\frac{25}{9}$$

Factor x^{2}-\frac{10}{3}x+\frac{25}{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}.

$$\left(x-\frac{5}{3}\right)^{2}=\frac{25}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{3}\right)^{2}}=\sqrt{\frac{25}{9}}$$

Simplify.

$$x-\frac{5}{3}=\frac{5}{3} x-\frac{5}{3}=-\frac{5}{3}$$

Add \frac{5}{3} to both sides of the equation.

$$x=\frac{10}{3} x=0$$