Solve (x+1/2)^2-(1/2-x)(1/2+x)=(x-3/2)(x+2)+6

$$( x + \frac { 1 } { 2 } ) ^ { 2 } - ( \frac { 1 } { 2 } - x ) ( \frac { 1 } { 2 } + x ) = ( x - \frac { 3 } { 2 } ) ( x + 2 ) + 6$$

Solve for x

$$x=-2$$

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

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+\frac{1}{2}\right)^{2}.

$$x^{2}+x+\frac{1}{4}-\left(\frac{1}{2}-x\right)\left(\frac{1}{2}+x\right)=\left(x-\frac{3}{2}\right)\left(x+2\right)+6$$

Consider \left(\frac{1}{2}-x\right)\left(\frac{1}{2}+x\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square \frac{1}{2}.

$$x^{2}+x+\frac{1}{4}-\left(\frac{1}{4}-x^{2}\right)=\left(x-\frac{3}{2}\right)\left(x+2\right)+6$$

To find the opposite of \frac{1}{4}-x^{2}, find the opposite of each term.

$$x^{2}+x+\frac{1}{4}-\frac{1}{4}+x^{2}=\left(x-\frac{3}{2}\right)\left(x+2\right)+6$$

Subtract \frac{1}{4} from \frac{1}{4} to get 0.

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

Combine x^{2} and x^{2} to get 2x^{2}.

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

Use the distributive property to multiply x-\frac{3}{2} by x+2 and combine like terms.

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

Add -3 and 6 to get 3.

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

Subtract x^{2} from both sides.

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

Combine 2x^{2} and -x^{2} to get x^{2}.

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

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

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

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

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

Subtract 3 from both sides.

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

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

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

Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.

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

Multiply -4 times -3.

$$x=\frac{-\frac{1}{2}±\sqrt{\frac{1}{4}+12}}{2}$$

Add \frac{1}{4} to 12.

$$x=\frac{-\frac{1}{2}±\sqrt{\frac{49}{4}}}{2}$$

Take the square root of \frac{49}{4}.

$$x=\frac{-\frac{1}{2}±\frac{7}{2}}{2}$$

Now solve the equation x=\frac{-\frac{1}{2}±\frac{7}{2}}{2} when ± is plus. Add -\frac{1}{2} to \frac{7}{2} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

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

Now solve the equation x=\frac{-\frac{1}{2}±\frac{7}{2}}{2} when ± is minus. Subtract \frac{7}{2} from -\frac{1}{2} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{4}{2}$$

Divide -4 by 2.

$$x=-2$$

The equation is now solved.

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

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+\frac{1}{2}\right)^{2}.

$$x^{2}+x+\frac{1}{4}-\left(\frac{1}{2}-x\right)\left(\frac{1}{2}+x\right)=\left(x-\frac{3}{2}\right)\left(x+2\right)+6$$

Consider \left(\frac{1}{2}-x\right)\left(\frac{1}{2}+x\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square \frac{1}{2}.

$$x^{2}+x+\frac{1}{4}-\left(\frac{1}{4}-x^{2}\right)=\left(x-\frac{3}{2}\right)\left(x+2\right)+6$$

To find the opposite of \frac{1}{4}-x^{2}, find the opposite of each term.

$$x^{2}+x+\frac{1}{4}-\frac{1}{4}+x^{2}=\left(x-\frac{3}{2}\right)\left(x+2\right)+6$$

Subtract \frac{1}{4} from \frac{1}{4} to get 0.

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

Combine x^{2} and x^{2} to get 2x^{2}.

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

Use the distributive property to multiply x-\frac{3}{2} by x+2 and combine like terms.

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

Add -3 and 6 to get 3.

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

Subtract x^{2} from both sides.

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

Combine 2x^{2} and -x^{2} to get x^{2}.

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

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

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

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

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

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.

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

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

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

Add 3 to \frac{1}{16}.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{49}{16}$$

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}.

$$\left(x+\frac{1}{4}\right)^{2}=\frac{49}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{4}\right)^{2}}=\sqrt{\frac{49}{16}}$$

Simplify.

$$x+\frac{1}{4}=\frac{7}{4} x+\frac{1}{4}=-\frac{7}{4}$$

Subtract \frac{1}{4} from both sides of the equation.

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