Solve x^2-3x=1/2(x-3)

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

Solve for x

$$x=\frac{1}{2}=0.5$$

$$x=3$$

Steps Using the Quadratic Formula

Use the distributive property to multiply \frac{1}{2} by x-3.

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

Subtract \frac{1}{2}x from both sides.

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

Combine -3x and -\frac{1}{2}x to get -\frac{7}{2}x.

$$x^{2}-\frac{7}{2}x=-\frac{3}{2}$$

Add \frac{3}{2} to both sides.

$$x^{2}-\frac{7}{2}x+\frac{3}{2}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -\frac{7}{2} for b, and \frac{3}{2} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-\frac{7}{2}\right)±\sqrt{\left(-\frac{7}{2}\right)^{2}-4\times \frac{3}{2}}}{2}$$

Square -\frac{7}{2} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\left(-\frac{7}{2}\right)±\sqrt{\frac{49}{4}-4\times \frac{3}{2}}}{2}$$

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

$$x=\frac{-\left(-\frac{7}{2}\right)±\sqrt{\frac{49}{4}-6}}{2}$$

Add \frac{49}{4} to -6.

$$x=\frac{-\left(-\frac{7}{2}\right)±\sqrt{\frac{25}{4}}}{2}$$

Take the square root of \frac{25}{4}.

$$x=\frac{-\left(-\frac{7}{2}\right)±\frac{5}{2}}{2}$$

The opposite of -\frac{7}{2} is \frac{7}{2}.

$$x=\frac{\frac{7}{2}±\frac{5}{2}}{2}$$

Now solve the equation x=\frac{\frac{7}{2}±\frac{5}{2}}{2} when ± is plus. Add \frac{7}{2} to \frac{5}{2} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{6}{2}$$

Divide 6 by 2.

$$x=3$$

Now solve the equation x=\frac{\frac{7}{2}±\frac{5}{2}}{2} when ± is minus. Subtract \frac{5}{2} from \frac{7}{2} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

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

The equation is now solved.

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

Steps for Completing the Square

Use the distributive property to multiply \frac{1}{2} by x-3.

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

Subtract \frac{1}{2}x from both sides.

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

Combine -3x and -\frac{1}{2}x to get -\frac{7}{2}x.

$$x^{2}-\frac{7}{2}x=-\frac{3}{2}$$

Divide -\frac{7}{2}, the coefficient of the x term, by 2 to get -\frac{7}{4}. Then add the square of -\frac{7}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{7}{2}x+\left(-\frac{7}{4}\right)^{2}=-\frac{3}{2}+\left(-\frac{7}{4}\right)^{2}$$

Square -\frac{7}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{7}{2}x+\frac{49}{16}=-\frac{3}{2}+\frac{49}{16}$$

Add -\frac{3}{2} to \frac{49}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{7}{2}x+\frac{49}{16}=\frac{25}{16}$$

Factor x^{2}-\frac{7}{2}x+\frac{49}{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}.

$$\left(x-\frac{7}{4}\right)^{2}=\frac{25}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{4}\right)^{2}}=\sqrt{\frac{25}{16}}$$

Simplify.

$$x-\frac{7}{4}=\frac{5}{4} x-\frac{7}{4}=-\frac{5}{4}$$

Add \frac{7}{4} to both sides of the equation.

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