Solve (x)=(x+3)^2+4

$$( x ) = ( x + 3 ) ^ { 2 } + 4$$

Solve for x (complex solution)

$$x=\frac{-3\sqrt{3}i-5}{2}\approx -2.5-2.598076211i$$

$$x=\frac{-5+3\sqrt{3}i}{2}\approx -2.5+2.598076211i$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+3\right)^{2}.

$$x=x^{2}+6x+9+4$$

Add 9 and 4 to get 13.

$$x=x^{2}+6x+13$$

Subtract x^{2} from both sides.

$$x-x^{2}=6x+13$$

Subtract 6x from both sides.

$$x-x^{2}-6x=13$$

Combine x and -6x to get -5x.

$$-5x-x^{2}=13$$

Subtract 13 from both sides.

$$-5x-x^{2}-13=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-x^{2}-5x-13=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -5 for b, and -13 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\left(-1\right)\left(-13\right)}}{2\left(-1\right)}$$

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\left(-1\right)\left(-13\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-5\right)±\sqrt{25+4\left(-13\right)}}{2\left(-1\right)}$$

Multiply 4 times -13.

$$x=\frac{-\left(-5\right)±\sqrt{25-52}}{2\left(-1\right)}$$

Add 25 to -52.

$$x=\frac{-\left(-5\right)±\sqrt{-27}}{2\left(-1\right)}$$

Take the square root of -27.

$$x=\frac{-\left(-5\right)±3\sqrt{3}i}{2\left(-1\right)}$$

The opposite of -5 is 5.

$$x=\frac{5±3\sqrt{3}i}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{5±3\sqrt{3}i}{-2}$$

Now solve the equation x=\frac{5±3\sqrt{3}i}{-2} when ± is plus. Add 5 to 3i\sqrt{3}.

$$x=\frac{5+3\sqrt{3}i}{-2}$$

Divide 5+3i\sqrt{3} by -2.

$$x=\frac{-3\sqrt{3}i-5}{2}$$

Now solve the equation x=\frac{5±3\sqrt{3}i}{-2} when ± is minus. Subtract 3i\sqrt{3} from 5.

$$x=\frac{-3\sqrt{3}i+5}{-2}$$

Divide 5-3i\sqrt{3} by -2.

$$x=\frac{-5+3\sqrt{3}i}{2}$$

The equation is now solved.

$$x=\frac{-3\sqrt{3}i-5}{2} x=\frac{-5+3\sqrt{3}i}{2}$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+3\right)^{2}.

$$x=x^{2}+6x+9+4$$

Add 9 and 4 to get 13.

$$x=x^{2}+6x+13$$

Subtract x^{2} from both sides.

$$x-x^{2}=6x+13$$

Subtract 6x from both sides.

$$x-x^{2}-6x=13$$

Combine x and -6x to get -5x.

$$-5x-x^{2}=13$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-x^{2}-5x=13$$

Divide both sides by -1.

$$\frac{-x^{2}-5x}{-1}=\frac{13}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\left(-\frac{5}{-1}\right)x=\frac{13}{-1}$$

Divide -5 by -1.

$$x^{2}+5x=\frac{13}{-1}$$

Divide 13 by -1.

$$x^{2}+5x=-13$$

Divide 5, the coefficient of the x term, by 2 to get \frac{5}{2}. Then add the square of \frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+5x+\left(\frac{5}{2}\right)^{2}=-13+\left(\frac{5}{2}\right)^{2}$$

Square \frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+5x+\frac{25}{4}=-13+\frac{25}{4}$$

Add -13 to \frac{25}{4}.

$$x^{2}+5x+\frac{25}{4}=-\frac{27}{4}$$

Factor x^{2}+5x+\frac{25}{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}.

$$\left(x+\frac{5}{2}\right)^{2}=-\frac{27}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{2}\right)^{2}}=\sqrt{-\frac{27}{4}}$$

Simplify.

$$x+\frac{5}{2}=\frac{3\sqrt{3}i}{2} x+\frac{5}{2}=-\frac{3\sqrt{3}i}{2}$$

Subtract \frac{5}{2} from both sides of the equation.

$$x=\frac{-5+3\sqrt{3}i}{2} x=\frac{-3\sqrt{3}i-5}{2}$$