Solve x^2+2x^2=5x

$$x ^ { 2 } + 2 x ^ { 2 } = 5 x$$

Solve for x

$$x = \frac{5}{3} = 1\frac{2}{3} \approx 1.666666667$$

$$x=0$$

Steps Using Factoring

Combine x^{2} and 2x^{2} to get 3x^{2}.

$$3x^{2}=5x$$

Subtract 5x from both sides.

$$3x^{2}-5x=0$$

Factor out x.

$$x\left(3x-5\right)=0$$

To find equation solutions, solve x=0 and 3x-5=0.

$$x=0 x=\frac{5}{3}$$

Steps Using the Quadratic Formula

Combine x^{2} and 2x^{2} to get 3x^{2}.

$$3x^{2}=5x$$

Subtract 5x from both sides.

$$3x^{2}-5x=0$$

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

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

Take the square root of \left(-5\right)^{2}.

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

The opposite of -5 is 5.

$$x=\frac{5±5}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{5±5}{6}$$

Now solve the equation x=\frac{5±5}{6} when ± is plus. Add 5 to 5.

$$x=\frac{10}{6}$$

Reduce the fraction \frac{10}{6} to lowest terms by extracting and canceling out 2.

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

Now solve the equation x=\frac{5±5}{6} when ± is minus. Subtract 5 from 5.

$$x=\frac{0}{6}$$

Divide 0 by 6.

$$x=0$$

The equation is now solved.

$$x=\frac{5}{3} x=0$$

Steps for Completing the Square

Combine x^{2} and 2x^{2} to get 3x^{2}.

$$3x^{2}=5x$$

Subtract 5x from both sides.

$$3x^{2}-5x=0$$

Divide both sides by 3.

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

Dividing by 3 undoes the multiplication by 3.

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

Divide 0 by 3.

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

Divide -\frac{5}{3}, the coefficient of the x term, by 2 to get -\frac{5}{6}. Then add the square of -\frac{5}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{5}{3}x+\left(-\frac{5}{6}\right)^{2}=\left(-\frac{5}{6}\right)^{2}$$

Square -\frac{5}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{5}{3}x+\frac{25}{36}=\frac{25}{36}$$

Factor x^{2}-\frac{5}{3}x+\frac{25}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{5}{6}\right)^{2}=\frac{25}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{6}\right)^{2}}=\sqrt{\frac{25}{36}}$$

Simplify.

$$x-\frac{5}{6}=\frac{5}{6} x-\frac{5}{6}=-\frac{5}{6}$$

Add \frac{5}{6} to both sides of the equation.

$$x=\frac{5}{3} x=0$$