Solve 6^2=x(x-3)

$${ 6 }^{ 2 } =x(x-3)$$

Solve for x

$$x = \frac{3 \sqrt{17} + 3}{2} \approx 7.684658438$$

$$x=\frac{3-3\sqrt{17}}{2}\approx -4.684658438$$

Steps Using the Quadratic Formula

Calculate 6 to the power of 2 and get 36.

$$36=x\left(x-3\right)$$

Use the distributive property to multiply x by x-3.

$$36=x^{2}-3x$$

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

$$x^{2}-3x=36$$

Subtract 36 from both sides.

$$x^{2}-3x-36=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -3 for b, and -36 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-36\right)}}{2}$$

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\left(-36\right)}}{2}$$

Multiply -4 times -36.

$$x=\frac{-\left(-3\right)±\sqrt{9+144}}{2}$$

Add 9 to 144.

$$x=\frac{-\left(-3\right)±\sqrt{153}}{2}$$

Take the square root of 153.

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

The opposite of -3 is 3.

$$x=\frac{3±3\sqrt{17}}{2}$$

Now solve the equation x=\frac{3±3\sqrt{17}}{2} when ± is plus. Add 3 to 3\sqrt{17}.

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

Now solve the equation x=\frac{3±3\sqrt{17}}{2} when ± is minus. Subtract 3\sqrt{17} from 3.

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

The equation is now solved.

$$x=\frac{3\sqrt{17}+3}{2} x=\frac{3-3\sqrt{17}}{2}$$

Steps for Completing the Square

Calculate 6 to the power of 2 and get 36.

$$36=x\left(x-3\right)$$

Use the distributive property to multiply x by x-3.

$$36=x^{2}-3x$$

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

$$x^{2}-3x=36$$

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

$$x^{2}-3x+\left(-\frac{3}{2}\right)^{2}=36+\left(-\frac{3}{2}\right)^{2}$$

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-3x+\frac{9}{4}=36+\frac{9}{4}$$

Add 36 to \frac{9}{4}.

$$x^{2}-3x+\frac{9}{4}=\frac{153}{4}$$

Factor x^{2}-3x+\frac{9}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{3}{2}\right)^{2}=\frac{153}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{\frac{153}{4}}$$

Simplify.

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

Add \frac{3}{2} to both sides of the equation.

$$x=\frac{3\sqrt{17}+3}{2} x=\frac{3-3\sqrt{17}}{2}$$