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

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

Solve for x

$$x=-1$$

$$x = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

Steps Using Factoring By Grouping

Multiply both sides of the equation by 2.

$$2x^{2}-x-3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^{2}+ax+bx-3. To find a and b, set up a system to be solved.

$$a+b=-1 ab=2\left(-3\right)=-6$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -6.

$$1,-6 2,-3$$

Calculate the sum for each pair.

$$1-6=-5 2-3=-1$$

The solution is the pair that gives sum -1.

$$a=-3 b=2$$

Rewrite 2x^{2}-x-3 as \left(2x^{2}-3x\right)+\left(2x-3\right).

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

Factor out x in 2x^{2}-3x.

$$x\left(2x-3\right)+2x-3$$

Factor out common term 2x-3 by using distributive property.

$$\left(2x-3\right)\left(x+1\right)$$

To find equation solutions, solve 2x-3=0 and x+1=0.

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

Steps Using the Quadratic Formula

Multiply both sides of the equation by 2.

$$2x^{2}-x-3=0$$

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

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

Multiply -4 times 2.

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

Multiply -8 times -3.

$$x=\frac{-\left(-1\right)±\sqrt{1+24}}{2\times 2}$$

Add 1 to 24.

$$x=\frac{-\left(-1\right)±\sqrt{25}}{2\times 2}$$

Take the square root of 25.

$$x=\frac{-\left(-1\right)±5}{2\times 2}$$

The opposite of -1 is 1.

$$x=\frac{1±5}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{1±5}{4}$$

Now solve the equation x=\frac{1±5}{4} when ± is plus. Add 1 to 5.

$$x=\frac{6}{4}$$

Reduce the fraction \frac{6}{4} to lowest terms by extracting and canceling out 2.

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

Now solve the equation x=\frac{1±5}{4} when ± is minus. Subtract 5 from 1.

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

Divide -4 by 4.

$$x=-1$$

The equation is now solved.

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

Steps for Completing the Square

Multiply both sides of the equation by 2.

$$2x^{2}-x-3=0$$

Add 3 to both sides. Anything plus zero gives itself.

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

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

Divide -\frac{1}{2}, the coefficient of the x term, by 2 to get -\frac{1}{4}. Then add the square of -\frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=\frac{3}{2}+\frac{1}{16}$$

Add \frac{3}{2} to \frac{1}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=\frac{25}{16}$$

Factor x^{2}-\frac{1}{2}x+\frac{1}{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}.

$$\left(x-\frac{1}{4}\right)^{2}=\frac{25}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{4}\right)^{2}}=\sqrt{\frac{25}{16}}$$

Simplify.

$$x-\frac{1}{4}=\frac{5}{4} x-\frac{1}{4}=-\frac{5}{4}$$

Add \frac{1}{4} to both sides of the equation.

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