Solve x^2-3x=3x^2-4x

$${ x }^{ 2 } -3x = 3 { x }^{ 2 } -4x$$

Solve for x

$$x=\frac{1}{2}=0.5$$

$$x=0$$

Steps Using Factoring

Subtract 3x^{2} from both sides.

$$x^{2}-3x-3x^{2}=-4x$$

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

$$-2x^{2}-3x=-4x$$

Add 4x to both sides.

$$-2x^{2}-3x+4x=0$$

Combine -3x and 4x to get x.

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

Factor out x.

$$x\left(-2x+1\right)=0$$

To find equation solutions, solve x=0 and -2x+1=0.

$$x=0 x=\frac{1}{2}$$

Steps Using the Quadratic Formula

Subtract 3x^{2} from both sides.

$$x^{2}-3x-3x^{2}=-4x$$

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

$$-2x^{2}-3x=-4x$$

Add 4x to both sides.

$$-2x^{2}-3x+4x=0$$

Combine -3x and 4x to get x.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, 1 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-1±\sqrt{1^{2}}}{2\left(-2\right)}$$

Take the square root of 1^{2}.

$$x=\frac{-1±1}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-1±1}{-4}$$

Now solve the equation x=\frac{-1±1}{-4} when ± is plus. Add -1 to 1.

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

Divide 0 by -4.

$$x=0$$

Now solve the equation x=\frac{-1±1}{-4} when ± is minus. Subtract 1 from -1.

$$x=-\frac{2}{-4}$$

Reduce the fraction \frac{-2}{-4} to lowest terms by extracting and canceling out 2.

$$x=\frac{1}{2}$$

The equation is now solved.

$$x=0 x=\frac{1}{2}$$

Steps for Completing the Square

Subtract 3x^{2} from both sides.

$$x^{2}-3x-3x^{2}=-4x$$

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

$$-2x^{2}-3x=-4x$$

Add 4x to both sides.

$$-2x^{2}-3x+4x=0$$

Combine -3x and 4x to get x.

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

Divide both sides by -2.

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

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\frac{1}{-2}x=\frac{0}{-2}$$

Divide 1 by -2.

$$x^{2}-\frac{1}{2}x=\frac{0}{-2}$$

Divide 0 by -2.

$$x^{2}-\frac{1}{2}x=0$$

Divide -\frac{1}{2}, the coefficient of the x term, by 2 to get -\frac{1}{4}. Then add the square of -\frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{2}x+\left(-\frac{1}{4}\right)^{2}=\left(-\frac{1}{4}\right)^{2}$$

Square -\frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=\frac{1}{16}$$

Factor x^{2}-\frac{1}{2}x+\frac{1}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{4}\right)^{2}=\frac{1}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{4}\right)^{2}}=\sqrt{\frac{1}{16}}$$

Simplify.

$$x-\frac{1}{4}=\frac{1}{4} x-\frac{1}{4}=-\frac{1}{4}$$

Add \frac{1}{4} to both sides of the equation.

$$x=\frac{1}{2} x=0$$