Solve 0=1/2x^2-4x+6

$$0 = \frac{ 1 }{ 2 } { x }^{ 2 } -4x+6$$

Solve for x

$$x=2$$

$$x=6$$

Steps Using the Quadratic Formula

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

$$\frac{1}{2}x^{2}-4x+6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{1}{2} for a, -4 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times \frac{1}{2}\times 6}}{2\times \frac{1}{2}}$$

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\times \frac{1}{2}\times 6}}{2\times \frac{1}{2}}$$

Multiply -4 times \frac{1}{2}.

$$x=\frac{-\left(-4\right)±\sqrt{16-2\times 6}}{2\times \frac{1}{2}}$$

Multiply -2 times 6.

$$x=\frac{-\left(-4\right)±\sqrt{16-12}}{2\times \frac{1}{2}}$$

Add 16 to -12.

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

Take the square root of 4.

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

The opposite of -4 is 4.

$$x=\frac{4±2}{2\times \frac{1}{2}}$$

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

$$x=\frac{4±2}{1}$$

Now solve the equation x=\frac{4±2}{1} when ± is plus. Add 4 to 2.

$$x=\frac{6}{1}$$

Divide 6 by 1.

$$x=6$$

Now solve the equation x=\frac{4±2}{1} when ± is minus. Subtract 2 from 4.

$$x=\frac{2}{1}$$

Divide 2 by 1.

$$x=2$$

The equation is now solved.

$$x=6 x=2$$

Steps for Completing the Square

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

$$\frac{1}{2}x^{2}-4x+6=0$$

Subtract 6 from both sides. Anything subtracted from zero gives its negation.

$$\frac{1}{2}x^{2}-4x=-6$$

Multiply both sides by 2.

$$\frac{\frac{1}{2}x^{2}-4x}{\frac{1}{2}}=-\frac{6}{\frac{1}{2}}$$

Dividing by \frac{1}{2} undoes the multiplication by \frac{1}{2}.

$$x^{2}+\left(-\frac{4}{\frac{1}{2}}\right)x=-\frac{6}{\frac{1}{2}}$$

Divide -4 by \frac{1}{2} by multiplying -4 by the reciprocal of \frac{1}{2}.

$$x^{2}-8x=-\frac{6}{\frac{1}{2}}$$

Divide -6 by \frac{1}{2} by multiplying -6 by the reciprocal of \frac{1}{2}.

$$x^{2}-8x=-12$$

Divide -8, the coefficient of the x term, by 2 to get -4. Then add the square of -4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-8x+\left(-4\right)^{2}=-12+\left(-4\right)^{2}$$

Square -4.

$$x^{2}-8x+16=-12+16$$

Add -12 to 16.

$$x^{2}-8x+16=4$$

Factor x^{2}-8x+16. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-4=2 x-4=-2$$

Add 4 to both sides of the equation.

$$x=6 x=2$$