Solve 15x-5x^2=x

$$15 x - 5 x ^ { 2 } = x$$

Solve for x

$$x = \frac{14}{5} = 2\frac{4}{5} = 2.8$$

$$x=0$$

Steps Using Factoring

Subtract x from both sides.

$$15x-5x^{2}-x=0$$

Combine 15x and -x to get 14x.

$$14x-5x^{2}=0$$

Factor out x.

$$x\left(14-5x\right)=0$$

To find equation solutions, solve x=0 and 14-5x=0.

$$x=0 x=\frac{14}{5}$$

Steps Using the Quadratic Formula

Subtract x from both sides.

$$15x-5x^{2}-x=0$$

Combine 15x and -x to get 14x.

$$14x-5x^{2}=0$$

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.

$$-5x^{2}+14x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -5 for a, 14 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-14±\sqrt{14^{2}}}{2\left(-5\right)}$$

Take the square root of 14^{2}.

$$x=\frac{-14±14}{2\left(-5\right)}$$

Multiply 2 times -5.

$$x=\frac{-14±14}{-10}$$

Now solve the equation x=\frac{-14±14}{-10} when ± is plus. Add -14 to 14.

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

Divide 0 by -10.

$$x=0$$

Now solve the equation x=\frac{-14±14}{-10} when ± is minus. Subtract 14 from -14.

$$x=-\frac{28}{-10}$$

Reduce the fraction \frac{-28}{-10} to lowest terms by extracting and canceling out 2.

$$x=\frac{14}{5}$$

The equation is now solved.

$$x=0 x=\frac{14}{5}$$

Steps for Completing the Square

Subtract x from both sides.

$$15x-5x^{2}-x=0$$

Combine 15x and -x to get 14x.

$$14x-5x^{2}=0$$

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.

$$-5x^{2}+14x=0$$

Divide both sides by -5.

$$\frac{-5x^{2}+14x}{-5}=\frac{0}{-5}$$

Dividing by -5 undoes the multiplication by -5.

$$x^{2}+\frac{14}{-5}x=\frac{0}{-5}$$

Divide 14 by -5.

$$x^{2}-\frac{14}{5}x=\frac{0}{-5}$$

Divide 0 by -5.

$$x^{2}-\frac{14}{5}x=0$$

Divide -\frac{14}{5}, the coefficient of the x term, by 2 to get -\frac{7}{5}. Then add the square of -\frac{7}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{14}{5}x+\left(-\frac{7}{5}\right)^{2}=\left(-\frac{7}{5}\right)^{2}$$

Square -\frac{7}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{14}{5}x+\frac{49}{25}=\frac{49}{25}$$

Factor x^{2}-\frac{14}{5}x+\frac{49}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{7}{5}\right)^{2}=\frac{49}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{5}\right)^{2}}=\sqrt{\frac{49}{25}}$$

Simplify.

$$x-\frac{7}{5}=\frac{7}{5} x-\frac{7}{5}=-\frac{7}{5}$$

Add \frac{7}{5} to both sides of the equation.

$$x=\frac{14}{5} x=0$$