Solve 6y^2+5y=9y-6

$$6 y ^ { 2 } + 5 y = 9 y - 6$$

Solve for y

$$y=\frac{1+2\sqrt{2}i}{3}\approx 0.333333333+0.942809042i$$

$$y=\frac{-2\sqrt{2}i+1}{3}\approx 0.333333333-0.942809042i$$

Steps Using the Quadratic Formula

Subtract 9y from both sides.

$$6y^{2}+5y-9y=-6$$

Combine 5y and -9y to get -4y.

$$6y^{2}-4y=-6$$

Add 6 to both sides.

$$6y^{2}-4y+6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, -4 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 6\times 6}}{2\times 6}$$

Square -4.

$$y=\frac{-\left(-4\right)±\sqrt{16-4\times 6\times 6}}{2\times 6}$$

Multiply -4 times 6.

$$y=\frac{-\left(-4\right)±\sqrt{16-24\times 6}}{2\times 6}$$

Multiply -24 times 6.

$$y=\frac{-\left(-4\right)±\sqrt{16-144}}{2\times 6}$$

Add 16 to -144.

$$y=\frac{-\left(-4\right)±\sqrt{-128}}{2\times 6}$$

Take the square root of -128.

$$y=\frac{-\left(-4\right)±8\sqrt{2}i}{2\times 6}$$

The opposite of -4 is 4.

$$y=\frac{4±8\sqrt{2}i}{2\times 6}$$

Multiply 2 times 6.

$$y=\frac{4±8\sqrt{2}i}{12}$$

Now solve the equation y=\frac{4±8\sqrt{2}i}{12} when ± is plus. Add 4 to 8i\sqrt{2}.

$$y=\frac{4+8\sqrt{2}i}{12}$$

Divide 4+8i\sqrt{2} by 12.

$$y=\frac{1+2\sqrt{2}i}{3}$$

Now solve the equation y=\frac{4±8\sqrt{2}i}{12} when ± is minus. Subtract 8i\sqrt{2} from 4.

$$y=\frac{-8\sqrt{2}i+4}{12}$$

Divide 4-8i\sqrt{2} by 12.

$$y=\frac{-2\sqrt{2}i+1}{3}$$

The equation is now solved.

$$y=\frac{1+2\sqrt{2}i}{3} y=\frac{-2\sqrt{2}i+1}{3}$$

Steps for Completing the Square

Subtract 9y from both sides.

$$6y^{2}+5y-9y=-6$$

Combine 5y and -9y to get -4y.

$$6y^{2}-4y=-6$$

Divide both sides by 6.

$$\frac{6y^{2}-4y}{6}=-\frac{6}{6}$$

Dividing by 6 undoes the multiplication by 6.

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

Reduce the fraction \frac{-4}{6} to lowest terms by extracting and canceling out 2.

$$y^{2}-\frac{2}{3}y=-\frac{6}{6}$$

Divide -6 by 6.

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

Divide -\frac{2}{3}, the coefficient of the x term, by 2 to get -\frac{1}{3}. Then add the square of -\frac{1}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{1}{3} by squaring both the numerator and the denominator of the fraction.

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

Add -1 to \frac{1}{9}.

$$y^{2}-\frac{2}{3}y+\frac{1}{9}=-\frac{8}{9}$$

Factor y^{2}-\frac{2}{3}y+\frac{1}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(y-\frac{1}{3}\right)^{2}=-\frac{8}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-\frac{1}{3}\right)^{2}}=\sqrt{-\frac{8}{9}}$$

Simplify.

$$y-\frac{1}{3}=\frac{2\sqrt{2}i}{3} y-\frac{1}{3}=-\frac{2\sqrt{2}i}{3}$$

Add \frac{1}{3} to both sides of the equation.

$$y=\frac{1+2\sqrt{2}i}{3} y=\frac{-2\sqrt{2}i+1}{3}$$