Solve 2Q-6=Q^2-6Q+9/Q

$$2 Q - 6 = \frac { Q ^ { 2 } - 6 Q + 9 } { Q }$$

Solve for Q

$$Q=-3$$

$$Q=3$$

Steps Using Difference of Squares

Variable Q cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by Q.

$$2QQ+Q\left(-6\right)=Q^{2}-6Q+9$$

Multiply Q and Q to get Q^{2}.

$$2Q^{2}+Q\left(-6\right)=Q^{2}-6Q+9$$

Subtract Q^{2} from both sides.

$$2Q^{2}+Q\left(-6\right)-Q^{2}=-6Q+9$$

Combine 2Q^{2} and -Q^{2} to get Q^{2}.

$$Q^{2}+Q\left(-6\right)=-6Q+9$$

Add 6Q to both sides.

$$Q^{2}+Q\left(-6\right)+6Q=9$$

Combine Q\left(-6\right) and 6Q to get 0.

$$Q^{2}=9$$

Subtract 9 from both sides.

$$Q^{2}-9=0$$

Consider Q^{2}-9. Rewrite Q^{2}-9 as Q^{2}-3^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(Q-3\right)\left(Q+3\right)=0$$

To find equation solutions, solve Q-3=0 and Q+3=0.

$$Q=3 Q=-3$$

Steps by Finding Square Root

Variable Q cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by Q.

$$2QQ+Q\left(-6\right)=Q^{2}-6Q+9$$

Multiply Q and Q to get Q^{2}.

$$2Q^{2}+Q\left(-6\right)=Q^{2}-6Q+9$$

Subtract Q^{2} from both sides.

$$2Q^{2}+Q\left(-6\right)-Q^{2}=-6Q+9$$

Combine 2Q^{2} and -Q^{2} to get Q^{2}.

$$Q^{2}+Q\left(-6\right)=-6Q+9$$

Add 6Q to both sides.

$$Q^{2}+Q\left(-6\right)+6Q=9$$

Combine Q\left(-6\right) and 6Q to get 0.

$$Q^{2}=9$$

Take the square root of both sides of the equation.

$$Q=3 Q=-3$$

Steps Using the Quadratic Formula

Variable Q cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by Q.

$$2QQ+Q\left(-6\right)=Q^{2}-6Q+9$$

Multiply Q and Q to get Q^{2}.

$$2Q^{2}+Q\left(-6\right)=Q^{2}-6Q+9$$

Subtract Q^{2} from both sides.

$$2Q^{2}+Q\left(-6\right)-Q^{2}=-6Q+9$$

Combine 2Q^{2} and -Q^{2} to get Q^{2}.

$$Q^{2}+Q\left(-6\right)=-6Q+9$$

Add 6Q to both sides.

$$Q^{2}+Q\left(-6\right)+6Q=9$$

Combine Q\left(-6\right) and 6Q to get 0.

$$Q^{2}=9$$

Subtract 9 from both sides.

$$Q^{2}-9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$Q=\frac{0±\sqrt{0^{2}-4\left(-9\right)}}{2}$$

Square 0.

$$Q=\frac{0±\sqrt{-4\left(-9\right)}}{2}$$

Multiply -4 times -9.

$$Q=\frac{0±\sqrt{36}}{2}$$

Take the square root of 36.

$$Q=\frac{0±6}{2}$$

Now solve the equation Q=\frac{0±6}{2} when ± is plus. Divide 6 by 2.

$$Q=3$$

Now solve the equation Q=\frac{0±6}{2} when ± is minus. Divide -6 by 2.

$$Q=-3$$

The equation is now solved.

$$Q=3 Q=-3$$