Solve 2x^2=1.4x

$$2 x ^ { 2 } = 1.4 x$$

Solve for x

$$x=0.7$$

$$x=0$$

Steps Using Factoring

Subtract 1.4x from both sides.

$$2x^{2}-1.4x=0$$

Factor out x.

$$x\left(2x-1.4\right)=0$$

To find equation solutions, solve x=0 and 2x-1.4=0.

$$x=0 x=\frac{7}{10}$$

Steps Using the Quadratic Formula

Subtract 1.4x from both sides.

$$2x^{2}-1.4x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -1.4 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-1.4\right)±\sqrt{\left(-1.4\right)^{2}}}{2\times 2}$$

Take the square root of \left(-1.4\right)^{2}.

$$x=\frac{-\left(-1.4\right)±\frac{7}{5}}{2\times 2}$$

The opposite of -1.4 is 1.4.

$$x=\frac{1.4±\frac{7}{5}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{1.4±\frac{7}{5}}{4}$$

Now solve the equation x=\frac{1.4±\frac{7}{5}}{4} when ± is plus. Add 1.4 to \frac{7}{5} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

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

Divide \frac{14}{5} by 4.

$$x=\frac{7}{10}$$

Now solve the equation x=\frac{1.4±\frac{7}{5}}{4} when ± is minus. Subtract \frac{7}{5} from 1.4 by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

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

Divide 0 by 4.

$$x=0$$

The equation is now solved.

$$x=\frac{7}{10} x=0$$

Steps for Completing the Square

Subtract 1.4x from both sides.

$$2x^{2}-1.4x=0$$

Divide both sides by 2.

$$\frac{2x^{2}-1.4x}{2}=\frac{0}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\left(-\frac{1.4}{2}\right)x=\frac{0}{2}$$

Divide -1.4 by 2.

$$x^{2}-0.7x=\frac{0}{2}$$

Divide 0 by 2.

$$x^{2}-0.7x=0$$

Divide -0.7, the coefficient of the x term, by 2 to get -0.35. Then add the square of -0.35 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-0.7x+\left(-0.35\right)^{2}=\left(-0.35\right)^{2}$$

Square -0.35 by squaring both the numerator and the denominator of the fraction.

$$x^{2}-0.7x+0.1225=0.1225$$

Factor x^{2}-0.7x+0.1225. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-0.35\right)^{2}=0.1225$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-0.35\right)^{2}}=\sqrt{0.1225}$$

Simplify.

$$x-0.35=\frac{7}{20} x-0.35=-\frac{7}{20}$$

Add 0.35 to both sides of the equation.

$$x=\frac{7}{10} x=0$$