Solve 1/2*(36+81+x^2)-18=x(9-x)

$$\frac { 1 } { 2 } \times ( 36 + 81 + x ^ { 2 } ) - 18 = x ( 9 - x )$$

Solve for x (complex solution)

$$x=3+3\sqrt{2}i\approx 3+4.242640687i$$

$$x=-3\sqrt{2}i+3\approx 3-4.242640687i$$

Steps Using the Quadratic Formula

Add 36 and 81 to get 117.

$$\frac{1}{2}\left(117+x^{2}\right)-18=x\left(9-x\right)$$

Use the distributive property to multiply \frac{1}{2} by 117+x^{2}.

$$\frac{117}{2}+\frac{1}{2}x^{2}-18=x\left(9-x\right)$$

Subtract 18 from \frac{117}{2} to get \frac{81}{2}.

$$\frac{81}{2}+\frac{1}{2}x^{2}=x\left(9-x\right)$$

Use the distributive property to multiply x by 9-x.

$$\frac{81}{2}+\frac{1}{2}x^{2}=9x-x^{2}$$

Subtract 9x from both sides.

$$\frac{81}{2}+\frac{1}{2}x^{2}-9x=-x^{2}$$

Add x^{2} to both sides.

$$\frac{81}{2}+\frac{1}{2}x^{2}-9x+x^{2}=0$$

Combine \frac{1}{2}x^{2} and x^{2} to get \frac{3}{2}x^{2}.

$$\frac{81}{2}+\frac{3}{2}x^{2}-9x=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$\frac{3}{2}x^{2}-9x+\frac{81}{2}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{3}{2} for a, -9 for b, and \frac{81}{2} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times \frac{3}{2}\times \frac{81}{2}}}{2\times \frac{3}{2}}$$

Square -9.

$$x=\frac{-\left(-9\right)±\sqrt{81-4\times \frac{3}{2}\times \frac{81}{2}}}{2\times \frac{3}{2}}$$

Multiply -4 times \frac{3}{2}.

$$x=\frac{-\left(-9\right)±\sqrt{81-6\times \frac{81}{2}}}{2\times \frac{3}{2}}$$

Multiply -6 times \frac{81}{2}.

$$x=\frac{-\left(-9\right)±\sqrt{81-243}}{2\times \frac{3}{2}}$$

Add 81 to -243.

$$x=\frac{-\left(-9\right)±\sqrt{-162}}{2\times \frac{3}{2}}$$

Take the square root of -162.

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

The opposite of -9 is 9.

$$x=\frac{9±9\sqrt{2}i}{2\times \frac{3}{2}}$$

Multiply 2 times \frac{3}{2}.

$$x=\frac{9±9\sqrt{2}i}{3}$$

Now solve the equation x=\frac{9±9\sqrt{2}i}{3} when ± is plus. Add 9 to 9i\sqrt{2}.

$$x=\frac{9+9\sqrt{2}i}{3}$$

Divide 9+9i\sqrt{2} by 3.

$$x=3+3\sqrt{2}i$$

Now solve the equation x=\frac{9±9\sqrt{2}i}{3} when ± is minus. Subtract 9i\sqrt{2} from 9.

$$x=\frac{-9\sqrt{2}i+9}{3}$$

Divide 9-9i\sqrt{2} by 3.

$$x=-3\sqrt{2}i+3$$

The equation is now solved.

$$x=3+3\sqrt{2}i x=-3\sqrt{2}i+3$$

Steps for Completing the Square

Add 36 and 81 to get 117.

$$\frac{1}{2}\left(117+x^{2}\right)-18=x\left(9-x\right)$$

Use the distributive property to multiply \frac{1}{2} by 117+x^{2}.

$$\frac{117}{2}+\frac{1}{2}x^{2}-18=x\left(9-x\right)$$

Subtract 18 from \frac{117}{2} to get \frac{81}{2}.

$$\frac{81}{2}+\frac{1}{2}x^{2}=x\left(9-x\right)$$

Use the distributive property to multiply x by 9-x.

$$\frac{81}{2}+\frac{1}{2}x^{2}=9x-x^{2}$$

Subtract 9x from both sides.

$$\frac{81}{2}+\frac{1}{2}x^{2}-9x=-x^{2}$$

Add x^{2} to both sides.

$$\frac{81}{2}+\frac{1}{2}x^{2}-9x+x^{2}=0$$

Combine \frac{1}{2}x^{2} and x^{2} to get \frac{3}{2}x^{2}.

$$\frac{81}{2}+\frac{3}{2}x^{2}-9x=0$$

Subtract \frac{81}{2} from both sides. Anything subtracted from zero gives its negation.

$$\frac{3}{2}x^{2}-9x=-\frac{81}{2}$$

Divide both sides of the equation by \frac{3}{2}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{3}{2}x^{2}-9x}{\frac{3}{2}}=-\frac{\frac{81}{2}}{\frac{3}{2}}$$

Dividing by \frac{3}{2} undoes the multiplication by \frac{3}{2}.

$$x^{2}+\left(-\frac{9}{\frac{3}{2}}\right)x=-\frac{\frac{81}{2}}{\frac{3}{2}}$$

Divide -9 by \frac{3}{2} by multiplying -9 by the reciprocal of \frac{3}{2}.

$$x^{2}-6x=-\frac{\frac{81}{2}}{\frac{3}{2}}$$

Divide -\frac{81}{2} by \frac{3}{2} by multiplying -\frac{81}{2} by the reciprocal of \frac{3}{2}.

$$x^{2}-6x=-27$$

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

$$x^{2}-6x+\left(-3\right)^{2}=-27+\left(-3\right)^{2}$$

Square -3.

$$x^{2}-6x+9=-27+9$$

Add -27 to 9.

$$x^{2}-6x+9=-18$$

Factor x^{2}-6x+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}.

$$\left(x-3\right)^{2}=-18$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-3\right)^{2}}=\sqrt{-18}$$

Simplify.

$$x-3=3\sqrt{2}i x-3=-3\sqrt{2}i$$

Add 3 to both sides of the equation.

$$x=3+3\sqrt{2}i x=-3\sqrt{2}i+3$$