Solve 1/3=3/2x^2-5/3x-4

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

Solve for x

$$x = \frac{\sqrt{259} + 5}{9} \approx 2.34371966$$

$$x=\frac{5-\sqrt{259}}{9}\approx -1.232608549$$

Steps Using the Quadratic Formula

Swap sides so that all variable terms are on the left hand side.

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

Subtract \frac{1}{3} from both sides.

$$\frac{3}{2}x^{2}-\frac{5}{3}x-4-\frac{1}{3}=0$$

Subtract \frac{1}{3} from -4 to get -\frac{13}{3}.

$$\frac{3}{2}x^{2}-\frac{5}{3}x-\frac{13}{3}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{3}{2} for a, -\frac{5}{3} for b, and -\frac{13}{3} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-\frac{5}{3}\right)±\sqrt{\left(-\frac{5}{3}\right)^{2}-4\times \frac{3}{2}\left(-\frac{13}{3}\right)}}{2\times \frac{3}{2}}$$

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

$$x=\frac{-\left(-\frac{5}{3}\right)±\sqrt{\frac{25}{9}-4\times \frac{3}{2}\left(-\frac{13}{3}\right)}}{2\times \frac{3}{2}}$$

Multiply -4 times \frac{3}{2}.

$$x=\frac{-\left(-\frac{5}{3}\right)±\sqrt{\frac{25}{9}-6\left(-\frac{13}{3}\right)}}{2\times \frac{3}{2}}$$

Multiply -6 times -\frac{13}{3}.

$$x=\frac{-\left(-\frac{5}{3}\right)±\sqrt{\frac{25}{9}+26}}{2\times \frac{3}{2}}$$

Add \frac{25}{9} to 26.

$$x=\frac{-\left(-\frac{5}{3}\right)±\sqrt{\frac{259}{9}}}{2\times \frac{3}{2}}$$

Take the square root of \frac{259}{9}.

$$x=\frac{-\left(-\frac{5}{3}\right)±\frac{\sqrt{259}}{3}}{2\times \frac{3}{2}}$$

The opposite of -\frac{5}{3} is \frac{5}{3}.

$$x=\frac{\frac{5}{3}±\frac{\sqrt{259}}{3}}{2\times \frac{3}{2}}$$

Multiply 2 times \frac{3}{2}.

$$x=\frac{\frac{5}{3}±\frac{\sqrt{259}}{3}}{3}$$

Now solve the equation x=\frac{\frac{5}{3}±\frac{\sqrt{259}}{3}}{3} when ± is plus. Add \frac{5}{3} to \frac{\sqrt{259}}{3}.

$$x=\frac{\sqrt{259}+5}{3\times 3}$$

Divide \frac{5+\sqrt{259}}{3} by 3.

$$x=\frac{\sqrt{259}+5}{9}$$

Now solve the equation x=\frac{\frac{5}{3}±\frac{\sqrt{259}}{3}}{3} when ± is minus. Subtract \frac{\sqrt{259}}{3} from \frac{5}{3}.

$$x=\frac{5-\sqrt{259}}{3\times 3}$$

Divide \frac{5-\sqrt{259}}{3} by 3.

$$x=\frac{5-\sqrt{259}}{9}$$

The equation is now solved.

$$x=\frac{\sqrt{259}+5}{9} x=\frac{5-\sqrt{259}}{9}$$

Steps for Completing the Square

Swap sides so that all variable terms are on the left hand side.

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

Add 4 to both sides.

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

Add \frac{1}{3} and 4 to get \frac{13}{3}.

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

Divide both sides of the equation by \frac{3}{2}, which is the same as multiplying both sides by the reciprocal of the fraction.

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

Dividing by \frac{3}{2} undoes the multiplication by \frac{3}{2}.

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

Divide -\frac{5}{3} by \frac{3}{2} by multiplying -\frac{5}{3} by the reciprocal of \frac{3}{2}.

$$x^{2}-\frac{10}{9}x=\frac{\frac{13}{3}}{\frac{3}{2}}$$

Divide \frac{13}{3} by \frac{3}{2} by multiplying \frac{13}{3} by the reciprocal of \frac{3}{2}.

$$x^{2}-\frac{10}{9}x=\frac{26}{9}$$

Divide -\frac{10}{9}, the coefficient of the x term, by 2 to get -\frac{5}{9}. Then add the square of -\frac{5}{9} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{10}{9}x+\left(-\frac{5}{9}\right)^{2}=\frac{26}{9}+\left(-\frac{5}{9}\right)^{2}$$

Square -\frac{5}{9} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{10}{9}x+\frac{25}{81}=\frac{26}{9}+\frac{25}{81}$$

Add \frac{26}{9} to \frac{25}{81} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{10}{9}x+\frac{25}{81}=\frac{259}{81}$$

Factor x^{2}-\frac{10}{9}x+\frac{25}{81}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{5}{9}\right)^{2}=\frac{259}{81}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{9}\right)^{2}}=\sqrt{\frac{259}{81}}$$

Simplify.

$$x-\frac{5}{9}=\frac{\sqrt{259}}{9} x-\frac{5}{9}=-\frac{\sqrt{259}}{9}$$

Add \frac{5}{9} to both sides of the equation.

$$x=\frac{\sqrt{259}+5}{9} x=\frac{5-\sqrt{259}}{9}$$