Solve left(x+2right)^2+2=1/2x+1

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

Solve for x (complex solution)

$$x=\frac{-7+\sqrt{31}i}{4}\approx -1.75+1.391941091i$$

$$x=\frac{-\sqrt{31}i-7}{4}\approx -1.75-1.391941091i$$

Steps Using the Quadratic Formula

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

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

Add 4 and 2 to get 6.

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

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

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

Combine 4x and -\frac{1}{2}x to get \frac{7}{2}x.

$$x^{2}+\frac{7}{2}x+6=1$$

Subtract 1 from both sides.

$$x^{2}+\frac{7}{2}x+6-1=0$$

Subtract 1 from 6 to get 5.

$$x^{2}+\frac{7}{2}x+5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, \frac{7}{2} for b, and 5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\frac{7}{2}±\sqrt{\left(\frac{7}{2}\right)^{2}-4\times 5}}{2}$$

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

$$x=\frac{-\frac{7}{2}±\sqrt{\frac{49}{4}-4\times 5}}{2}$$

Multiply -4 times 5.

$$x=\frac{-\frac{7}{2}±\sqrt{\frac{49}{4}-20}}{2}$$

Add \frac{49}{4} to -20.

$$x=\frac{-\frac{7}{2}±\sqrt{-\frac{31}{4}}}{2}$$

Take the square root of -\frac{31}{4}.

$$x=\frac{-\frac{7}{2}±\frac{\sqrt{31}i}{2}}{2}$$

Now solve the equation x=\frac{-\frac{7}{2}±\frac{\sqrt{31}i}{2}}{2} when ± is plus. Add -\frac{7}{2} to \frac{i\sqrt{31}}{2}.

$$x=\frac{-7+\sqrt{31}i}{2\times 2}$$

Divide \frac{-7+i\sqrt{31}}{2} by 2.

$$x=\frac{-7+\sqrt{31}i}{4}$$

Now solve the equation x=\frac{-\frac{7}{2}±\frac{\sqrt{31}i}{2}}{2} when ± is minus. Subtract \frac{i\sqrt{31}}{2} from -\frac{7}{2}.

$$x=\frac{-\sqrt{31}i-7}{2\times 2}$$

Divide \frac{-7-i\sqrt{31}}{2} by 2.

$$x=\frac{-\sqrt{31}i-7}{4}$$

The equation is now solved.

$$x=\frac{-7+\sqrt{31}i}{4} x=\frac{-\sqrt{31}i-7}{4}$$

Steps for Completing the Square

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

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

Add 4 and 2 to get 6.

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

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

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

Combine 4x and -\frac{1}{2}x to get \frac{7}{2}x.

$$x^{2}+\frac{7}{2}x+6=1$$

Subtract 6 from both sides.

$$x^{2}+\frac{7}{2}x=1-6$$

Subtract 6 from 1 to get -5.

$$x^{2}+\frac{7}{2}x=-5$$

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.

$$x^{2}+\frac{7}{2}x+\left(\frac{7}{4}\right)^{2}=-5+\left(\frac{7}{4}\right)^{2}$$

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

$$x^{2}+\frac{7}{2}x+\frac{49}{16}=-5+\frac{49}{16}$$

Add -5 to \frac{49}{16}.

$$x^{2}+\frac{7}{2}x+\frac{49}{16}=-\frac{31}{16}$$

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}.

$$\left(x+\frac{7}{4}\right)^{2}=-\frac{31}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{7}{4}\right)^{2}}=\sqrt{-\frac{31}{16}}$$

Simplify.

$$x+\frac{7}{4}=\frac{\sqrt{31}i}{4} x+\frac{7}{4}=-\frac{\sqrt{31}i}{4}$$

Subtract \frac{7}{4} from both sides of the equation.

$$x=\frac{-7+\sqrt{31}i}{4} x=\frac{-\sqrt{31}i-7}{4}$$