Solve 10x^2=8x

$$10 { x }^{ 2 } =8x$$

Solve for x

$$x=\frac{4}{5}=0.8$$

$$x=0$$

Steps Using Factoring

Subtract 8x from both sides.

$$10x^{2}-8x=0$$

Factor out x.

$$x\left(10x-8\right)=0$$

To find equation solutions, solve x=0 and 10x-8=0.

$$x=0 x=\frac{4}{5}$$

Steps Using the Quadratic Formula

Subtract 8x from both sides.

$$10x^{2}-8x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 10 for a, -8 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}}}{2\times 10}$$

Take the square root of \left(-8\right)^{2}.

$$x=\frac{-\left(-8\right)±8}{2\times 10}$$

The opposite of -8 is 8.

$$x=\frac{8±8}{2\times 10}$$

Multiply 2 times 10.

$$x=\frac{8±8}{20}$$

Now solve the equation x=\frac{8±8}{20} when ± is plus. Add 8 to 8.

$$x=\frac{16}{20}$$

Reduce the fraction \frac{16}{20} to lowest terms by extracting and canceling out 4.

$$x=\frac{4}{5}$$

Now solve the equation x=\frac{8±8}{20} when ± is minus. Subtract 8 from 8.

$$x=\frac{0}{20}$$

Divide 0 by 20.

$$x=0$$

The equation is now solved.

$$x=\frac{4}{5} x=0$$

Steps for Completing the Square

Subtract 8x from both sides.

$$10x^{2}-8x=0$$

Divide both sides by 10.

$$\frac{10x^{2}-8x}{10}=\frac{0}{10}$$

Dividing by 10 undoes the multiplication by 10.

$$x^{2}+\left(-\frac{8}{10}\right)x=\frac{0}{10}$$

Reduce the fraction \frac{-8}{10} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{4}{5}x=\frac{0}{10}$$

Divide 0 by 10.

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

Divide -\frac{4}{5}, the coefficient of the x term, by 2 to get -\frac{2}{5}. Then add the square of -\frac{2}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{2}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{4}{5}x+\frac{4}{25}=\frac{4}{25}$$

Factor x^{2}-\frac{4}{5}x+\frac{4}{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}.

$$\left(x-\frac{2}{5}\right)^{2}=\frac{4}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{2}{5}\right)^{2}}=\sqrt{\frac{4}{25}}$$

Simplify.

$$x-\frac{2}{5}=\frac{2}{5} x-\frac{2}{5}=-\frac{2}{5}$$

Add \frac{2}{5} to both sides of the equation.

$$x=\frac{4}{5} x=0$$