Solve (x)=x^2-x/3x+1

$$(x)= \frac{ { x }^{ 2 } -x }{ 3x+1 }$$

Solve for x

$$x=-1$$

$$x=0$$

Steps Using the Quadratic Formula

Subtract \frac{x^{2}-x}{3x+1} from both sides.

$$x-\frac{x^{2}-x}{3x+1}=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x times \frac{3x+1}{3x+1}.

$$\frac{x\left(3x+1\right)}{3x+1}-\frac{x^{2}-x}{3x+1}=0$$

Since \frac{x\left(3x+1\right)}{3x+1} and \frac{x^{2}-x}{3x+1} have the same denominator, subtract them by subtracting their numerators.

$$\frac{x\left(3x+1\right)-\left(x^{2}-x\right)}{3x+1}=0$$

Do the multiplications in x\left(3x+1\right)-\left(x^{2}-x\right).

$$\frac{3x^{2}+x-x^{2}+x}{3x+1}=0$$

Combine like terms in 3x^{2}+x-x^{2}+x.

$$\frac{2x^{2}+2x}{3x+1}=0$$

Variable x cannot be equal to -\frac{1}{3} since division by zero is not defined. Multiply both sides of the equation by 3x+1.

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

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

$$x=\frac{-2±\sqrt{2^{2}}}{2\times 2}$$

Take the square root of 2^{2}.

$$x=\frac{-2±2}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-2±2}{4}$$

Now solve the equation x=\frac{-2±2}{4} when ± is plus. Add -2 to 2.

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

Divide 0 by 4.

$$x=0$$

Now solve the equation x=\frac{-2±2}{4} when ± is minus. Subtract 2 from -2.

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

Divide -4 by 4.

$$x=-1$$

The equation is now solved.

$$x=0 x=-1$$

Steps for Completing the Square

Subtract \frac{x^{2}-x}{3x+1} from both sides.

$$x-\frac{x^{2}-x}{3x+1}=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x times \frac{3x+1}{3x+1}.

$$\frac{x\left(3x+1\right)}{3x+1}-\frac{x^{2}-x}{3x+1}=0$$

Since \frac{x\left(3x+1\right)}{3x+1} and \frac{x^{2}-x}{3x+1} have the same denominator, subtract them by subtracting their numerators.

$$\frac{x\left(3x+1\right)-\left(x^{2}-x\right)}{3x+1}=0$$

Do the multiplications in x\left(3x+1\right)-\left(x^{2}-x\right).

$$\frac{3x^{2}+x-x^{2}+x}{3x+1}=0$$

Combine like terms in 3x^{2}+x-x^{2}+x.

$$\frac{2x^{2}+2x}{3x+1}=0$$

Variable x cannot be equal to -\frac{1}{3} since division by zero is not defined. Multiply both sides of the equation by 3x+1.

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

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

Divide 2 by 2.

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

Divide 0 by 2.

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

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

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

Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+x+\frac{1}{4}=\frac{1}{4}$$

Factor x^{2}+x+\frac{1}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Subtract \frac{1}{2} from both sides of the equation.

$$x=0 x=-1$$