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

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

Solve for x

$$x = -\frac{4}{3} = -1\frac{1}{3} \approx -1.333333333$$

$$x=3$$

Steps Using the Quadratic Formula

Subtract x^{2} from both sides.

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

Add 2x to both sides.

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

Combine -\frac{1}{3}x and 2x to get \frac{5}{3}x.

$$\frac{5}{3}x+1-x^{2}=-3$$

Add 3 to both sides.

$$\frac{5}{3}x+1-x^{2}+3=0$$

Add 1 and 3 to get 4.

$$\frac{5}{3}x+4-x^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-x^{2}+\frac{5}{3}x+4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, \frac{5}{3} for b, and 4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\frac{5}{3}±\sqrt{\left(\frac{5}{3}\right)^{2}-4\left(-1\right)\times 4}}{2\left(-1\right)}$$

Square \frac{5}{3} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\frac{5}{3}±\sqrt{\frac{25}{9}-4\left(-1\right)\times 4}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\frac{5}{3}±\sqrt{\frac{25}{9}+4\times 4}}{2\left(-1\right)}$$

Multiply 4 times 4.

$$x=\frac{-\frac{5}{3}±\sqrt{\frac{25}{9}+16}}{2\left(-1\right)}$$

Add \frac{25}{9} to 16.

$$x=\frac{-\frac{5}{3}±\sqrt{\frac{169}{9}}}{2\left(-1\right)}$$

Take the square root of \frac{169}{9}.

$$x=\frac{-\frac{5}{3}±\frac{13}{3}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-\frac{5}{3}±\frac{13}{3}}{-2}$$

Now solve the equation x=\frac{-\frac{5}{3}±\frac{13}{3}}{-2} when ± is plus. Add -\frac{5}{3} to \frac{13}{3} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{\frac{8}{3}}{-2}$$

Divide \frac{8}{3} by -2.

$$x=-\frac{4}{3}$$

Now solve the equation x=\frac{-\frac{5}{3}±\frac{13}{3}}{-2} when ± is minus. Subtract \frac{13}{3} from -\frac{5}{3} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{6}{-2}$$

Divide -6 by -2.

$$x=3$$

The equation is now solved.

$$x=-\frac{4}{3} x=3$$

Steps for Completing the Square

Subtract x^{2} from both sides.

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

Add 2x to both sides.

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

Combine -\frac{1}{3}x and 2x to get \frac{5}{3}x.

$$\frac{5}{3}x+1-x^{2}=-3$$

Subtract 1 from both sides.

$$\frac{5}{3}x-x^{2}=-3-1$$

Subtract 1 from -3 to get -4.

$$\frac{5}{3}x-x^{2}=-4$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-x^{2}+\frac{5}{3}x=-4$$

Divide both sides by -1.

$$\frac{-x^{2}+\frac{5}{3}x}{-1}=-\frac{4}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{\frac{5}{3}}{-1}x=-\frac{4}{-1}$$

Divide \frac{5}{3} by -1.

$$x^{2}-\frac{5}{3}x=-\frac{4}{-1}$$

Divide -4 by -1.

$$x^{2}-\frac{5}{3}x=4$$

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.

$$x^{2}-\frac{5}{3}x+\left(-\frac{5}{6}\right)^{2}=4+\left(-\frac{5}{6}\right)^{2}$$

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

$$x^{2}-\frac{5}{3}x+\frac{25}{36}=4+\frac{25}{36}$$

Add 4 to \frac{25}{36}.

$$x^{2}-\frac{5}{3}x+\frac{25}{36}=\frac{169}{36}$$

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}.

$$\left(x-\frac{5}{6}\right)^{2}=\frac{169}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{6}\right)^{2}}=\sqrt{\frac{169}{36}}$$

Simplify.

$$x-\frac{5}{6}=\frac{13}{6} x-\frac{5}{6}=-\frac{13}{6}$$

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

$$x=3 x=-\frac{4}{3}$$