Solve x^2=17/3x+2

$${ x }^{ 2 } = \frac{ 17 }{ 3 } x+2$$

Solve for x

$$x=-\frac{1}{3}\approx -0.333333333$$

$$x=6$$

Steps Using the Quadratic Formula

Subtract \frac{17}{3}x from both sides.

$$x^{2}-\frac{17}{3}x=2$$

Subtract 2 from both sides.

$$x^{2}-\frac{17}{3}x-2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -\frac{17}{3} for b, and -2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-\frac{17}{3}\right)±\sqrt{\left(-\frac{17}{3}\right)^{2}-4\left(-2\right)}}{2}$$

Square -\frac{17}{3} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\left(-\frac{17}{3}\right)±\sqrt{\frac{289}{9}-4\left(-2\right)}}{2}$$

Multiply -4 times -2.

$$x=\frac{-\left(-\frac{17}{3}\right)±\sqrt{\frac{289}{9}+8}}{2}$$

Add \frac{289}{9} to 8.

$$x=\frac{-\left(-\frac{17}{3}\right)±\sqrt{\frac{361}{9}}}{2}$$

Take the square root of \frac{361}{9}.

$$x=\frac{-\left(-\frac{17}{3}\right)±\frac{19}{3}}{2}$$

The opposite of -\frac{17}{3} is \frac{17}{3}.

$$x=\frac{\frac{17}{3}±\frac{19}{3}}{2}$$

Now solve the equation x=\frac{\frac{17}{3}±\frac{19}{3}}{2} when ± is plus. Add \frac{17}{3} to \frac{19}{3} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{12}{2}$$

Divide 12 by 2.

$$x=6$$

Now solve the equation x=\frac{\frac{17}{3}±\frac{19}{3}}{2} when ± is minus. Subtract \frac{19}{3} from \frac{17}{3} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

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

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

$$x=-\frac{1}{3}$$

The equation is now solved.

$$x=6 x=-\frac{1}{3}$$

Steps for Completing the Square

Subtract \frac{17}{3}x from both sides.

$$x^{2}-\frac{17}{3}x=2$$

Divide -\frac{17}{3}, the coefficient of the x term, by 2 to get -\frac{17}{6}. Then add the square of -\frac{17}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{17}{3}x+\left(-\frac{17}{6}\right)^{2}=2+\left(-\frac{17}{6}\right)^{2}$$

Square -\frac{17}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{17}{3}x+\frac{289}{36}=2+\frac{289}{36}$$

Add 2 to \frac{289}{36}.

$$x^{2}-\frac{17}{3}x+\frac{289}{36}=\frac{361}{36}$$

Factor x^{2}-\frac{17}{3}x+\frac{289}{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}.

$$\left(x-\frac{17}{6}\right)^{2}=\frac{361}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{17}{6}\right)^{2}}=\sqrt{\frac{361}{36}}$$

Simplify.

$$x-\frac{17}{6}=\frac{19}{6} x-\frac{17}{6}=-\frac{19}{6}$$

Add \frac{17}{6} to both sides of the equation.

$$x=6 x=-\frac{1}{3}$$