Solve 9y^2=11y+3

$$9 y ^ { 2 } = 11 y + 3$$

Solve for y

$$y = \frac{\sqrt{229} + 11}{18} \approx 1.451819219$$

$$y=\frac{11-\sqrt{229}}{18}\approx -0.229596997$$

Steps Using the Quadratic Formula

Subtract 11y from both sides.

$$9y^{2}-11y=3$$

Subtract 3 from both sides.

$$9y^{2}-11y-3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, -11 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\times 9\left(-3\right)}}{2\times 9}$$

Square -11.

$$y=\frac{-\left(-11\right)±\sqrt{121-4\times 9\left(-3\right)}}{2\times 9}$$

Multiply -4 times 9.

$$y=\frac{-\left(-11\right)±\sqrt{121-36\left(-3\right)}}{2\times 9}$$

Multiply -36 times -3.

$$y=\frac{-\left(-11\right)±\sqrt{121+108}}{2\times 9}$$

Add 121 to 108.

$$y=\frac{-\left(-11\right)±\sqrt{229}}{2\times 9}$$

The opposite of -11 is 11.

$$y=\frac{11±\sqrt{229}}{2\times 9}$$

Multiply 2 times 9.

$$y=\frac{11±\sqrt{229}}{18}$$

Now solve the equation y=\frac{11±\sqrt{229}}{18} when ± is plus. Add 11 to \sqrt{229}.

$$y=\frac{\sqrt{229}+11}{18}$$

Now solve the equation y=\frac{11±\sqrt{229}}{18} when ± is minus. Subtract \sqrt{229} from 11.

$$y=\frac{11-\sqrt{229}}{18}$$

The equation is now solved.

$$y=\frac{\sqrt{229}+11}{18} y=\frac{11-\sqrt{229}}{18}$$

Steps for Completing the Square

Subtract 11y from both sides.

$$9y^{2}-11y=3$$

Divide both sides by 9.

$$\frac{9y^{2}-11y}{9}=\frac{3}{9}$$

Dividing by 9 undoes the multiplication by 9.

$$y^{2}-\frac{11}{9}y=\frac{3}{9}$$

Reduce the fraction \frac{3}{9} to lowest terms by extracting and canceling out 3.

$$y^{2}-\frac{11}{9}y=\frac{1}{3}$$

Divide -\frac{11}{9}, the coefficient of the x term, by 2 to get -\frac{11}{18}. Then add the square of -\frac{11}{18} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}-\frac{11}{9}y+\left(-\frac{11}{18}\right)^{2}=\frac{1}{3}+\left(-\frac{11}{18}\right)^{2}$$

Square -\frac{11}{18} by squaring both the numerator and the denominator of the fraction.

$$y^{2}-\frac{11}{9}y+\frac{121}{324}=\frac{1}{3}+\frac{121}{324}$$

Add \frac{1}{3} to \frac{121}{324} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y^{2}-\frac{11}{9}y+\frac{121}{324}=\frac{229}{324}$$

Factor y^{2}-\frac{11}{9}y+\frac{121}{324}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(y-\frac{11}{18}\right)^{2}=\frac{229}{324}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-\frac{11}{18}\right)^{2}}=\sqrt{\frac{229}{324}}$$

Simplify.

$$y-\frac{11}{18}=\frac{\sqrt{229}}{18} y-\frac{11}{18}=-\frac{\sqrt{229}}{18}$$

Add \frac{11}{18} to both sides of the equation.

$$y=\frac{\sqrt{229}+11}{18} y=\frac{11-\sqrt{229}}{18}$$