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

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

Solve for x

$$x=1$$

$$x=2$$

Steps Using the Quadratic Formula

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

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

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

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

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

$$x=\frac{-\left(-\frac{3}{2}\right)±\sqrt{\frac{9}{4}-4\times \frac{1}{2}}}{2\times \frac{1}{2}}$$

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

$$x=\frac{-\left(-\frac{3}{2}\right)±\sqrt{\frac{9}{4}-2}}{2\times \frac{1}{2}}$$

Add \frac{9}{4} to -2.

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

Take the square root of \frac{1}{4}.

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

The opposite of -\frac{3}{2} is \frac{3}{2}.

$$x=\frac{\frac{3}{2}±\frac{1}{2}}{2\times \frac{1}{2}}$$

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

$$x=\frac{\frac{3}{2}±\frac{1}{2}}{1}$$

Now solve the equation x=\frac{\frac{3}{2}±\frac{1}{2}}{1} when ± is plus. Add \frac{3}{2} to \frac{1}{2} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

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

Divide 2 by 1.

$$x=2$$

Now solve the equation x=\frac{\frac{3}{2}±\frac{1}{2}}{1} when ± is minus. Subtract \frac{1}{2} from \frac{3}{2} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

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

Divide 1 by 1.

$$x=1$$

The equation is now solved.

$$x=2 x=1$$

Steps for Completing the Square

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

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

Subtract 1 from both sides. Anything subtracted from zero gives its negation.

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

Multiply both sides by 2.

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

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

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

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

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

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

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

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.

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

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

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

Add -2 to \frac{9}{4}.

$$x^{2}-3x+\frac{9}{4}=\frac{1}{4}$$

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

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

$$x=2 x=1$$