Solve 0=35-2x+1/3x^2

$$0 = 35-2x+ \frac{ 1 }{ 3 } { x }^{ 2 }$$

Solve for x (complex solution)

$$x=3+4\sqrt{6}i\approx 3+9.797958971i$$

$$x=-4\sqrt{6}i+3\approx 3-9.797958971i$$

Steps Using the Quadratic Formula

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

$$35-2x+\frac{1}{3}x^{2}=0$$

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.

$$\frac{1}{3}x^{2}-2x+35=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{1}{3} for a, -2 for b, and 35 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times \frac{1}{3}\times 35}}{2\times \frac{1}{3}}$$

Square -2.

$$x=\frac{-\left(-2\right)±\sqrt{4-4\times \frac{1}{3}\times 35}}{2\times \frac{1}{3}}$$

Multiply -4 times \frac{1}{3}.

$$x=\frac{-\left(-2\right)±\sqrt{4-\frac{4}{3}\times 35}}{2\times \frac{1}{3}}$$

Multiply -\frac{4}{3} times 35.

$$x=\frac{-\left(-2\right)±\sqrt{4-\frac{140}{3}}}{2\times \frac{1}{3}}$$

Add 4 to -\frac{140}{3}.

$$x=\frac{-\left(-2\right)±\sqrt{-\frac{128}{3}}}{2\times \frac{1}{3}}$$

Take the square root of -\frac{128}{3}.

$$x=\frac{-\left(-2\right)±\frac{8\sqrt{6}i}{3}}{2\times \frac{1}{3}}$$

The opposite of -2 is 2.

$$x=\frac{2±\frac{8\sqrt{6}i}{3}}{2\times \frac{1}{3}}$$

Multiply 2 times \frac{1}{3}.

$$x=\frac{2±\frac{8\sqrt{6}i}{3}}{\frac{2}{3}}$$

Now solve the equation x=\frac{2±\frac{8\sqrt{6}i}{3}}{\frac{2}{3}} when ± is plus. Add 2 to \frac{8i\sqrt{6}}{3}.

$$x=\frac{\frac{8\sqrt{6}i}{3}+2}{\frac{2}{3}}$$

Divide 2+\frac{8i\sqrt{6}}{3} by \frac{2}{3} by multiplying 2+\frac{8i\sqrt{6}}{3} by the reciprocal of \frac{2}{3}.

$$x=3+4\sqrt{6}i$$

Now solve the equation x=\frac{2±\frac{8\sqrt{6}i}{3}}{\frac{2}{3}} when ± is minus. Subtract \frac{8i\sqrt{6}}{3} from 2.

$$x=\frac{-\frac{8\sqrt{6}i}{3}+2}{\frac{2}{3}}$$

Divide 2-\frac{8i\sqrt{6}}{3} by \frac{2}{3} by multiplying 2-\frac{8i\sqrt{6}}{3} by the reciprocal of \frac{2}{3}.

$$x=-4\sqrt{6}i+3$$

The equation is now solved.

$$x=3+4\sqrt{6}i x=-4\sqrt{6}i+3$$

Steps for Completing the Square

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

$$35-2x+\frac{1}{3}x^{2}=0$$

Subtract 35 from both sides. Anything subtracted from zero gives its negation.

$$-2x+\frac{1}{3}x^{2}=-35$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$\frac{1}{3}x^{2}-2x=-35$$

Multiply both sides by 3.

$$\frac{\frac{1}{3}x^{2}-2x}{\frac{1}{3}}=-\frac{35}{\frac{1}{3}}$$

Dividing by \frac{1}{3} undoes the multiplication by \frac{1}{3}.

$$x^{2}+\left(-\frac{2}{\frac{1}{3}}\right)x=-\frac{35}{\frac{1}{3}}$$

Divide -2 by \frac{1}{3} by multiplying -2 by the reciprocal of \frac{1}{3}.

$$x^{2}-6x=-\frac{35}{\frac{1}{3}}$$

Divide -35 by \frac{1}{3} by multiplying -35 by the reciprocal of \frac{1}{3}.

$$x^{2}-6x=-105$$

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.

$$x^{2}-6x+\left(-3\right)^{2}=-105+\left(-3\right)^{2}$$

Square -3.

$$x^{2}-6x+9=-105+9$$

Add -105 to 9.

$$x^{2}-6x+9=-96$$

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}.

$$\left(x-3\right)^{2}=-96$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-3\right)^{2}}=\sqrt{-96}$$

Simplify.

$$x-3=4\sqrt{6}i x-3=-4\sqrt{6}i$$

Add 3 to both sides of the equation.

$$x=3+4\sqrt{6}i x=-4\sqrt{6}i+3$$