Solve (x-3)^2+(x+4)^2=16

$$( x - 3 ) ^ { 2 } + ( x + 4 ) ^ { 2 } = 16$$

Solve for x (complex solution)

$$x=\frac{-1+\sqrt{17}i}{2}\approx -0.5+2.061552813i$$

$$x=\frac{-\sqrt{17}i-1}{2}\approx -0.5-2.061552813i$$

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}.

$$x^{2}-6x+9+\left(x+4\right)^{2}=16$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+4\right)^{2}.

$$x^{2}-6x+9+x^{2}+8x+16=16$$

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

$$2x^{2}-6x+9+8x+16=16$$

Combine -6x and 8x to get 2x.

$$2x^{2}+2x+9+16=16$$

Add 9 and 16 to get 25.

$$2x^{2}+2x+25=16$$

Subtract 16 from both sides.

$$2x^{2}+2x+25-16=0$$

Subtract 16 from 25 to get 9.

$$2x^{2}+2x+9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 2 for b, and 9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-2±\sqrt{2^{2}-4\times 2\times 9}}{2\times 2}$$

Square 2.

$$x=\frac{-2±\sqrt{4-4\times 2\times 9}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-2±\sqrt{4-8\times 9}}{2\times 2}$$

Multiply -8 times 9.

$$x=\frac{-2±\sqrt{4-72}}{2\times 2}$$

Add 4 to -72.

$$x=\frac{-2±\sqrt{-68}}{2\times 2}$$

Take the square root of -68.

$$x=\frac{-2±2\sqrt{17}i}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-2±2\sqrt{17}i}{4}$$

Now solve the equation x=\frac{-2±2\sqrt{17}i}{4} when ± is plus. Add -2 to 2i\sqrt{17}.

$$x=\frac{-2+2\sqrt{17}i}{4}$$

Divide -2+2i\sqrt{17} by 4.

$$x=\frac{-1+\sqrt{17}i}{2}$$

Now solve the equation x=\frac{-2±2\sqrt{17}i}{4} when ± is minus. Subtract 2i\sqrt{17} from -2.

$$x=\frac{-2\sqrt{17}i-2}{4}$$

Divide -2-2i\sqrt{17} by 4.

$$x=\frac{-\sqrt{17}i-1}{2}$$

The equation is now solved.

$$x=\frac{-1+\sqrt{17}i}{2} x=\frac{-\sqrt{17}i-1}{2}$$

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}.

$$x^{2}-6x+9+\left(x+4\right)^{2}=16$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+4\right)^{2}.

$$x^{2}-6x+9+x^{2}+8x+16=16$$

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

$$2x^{2}-6x+9+8x+16=16$$

Combine -6x and 8x to get 2x.

$$2x^{2}+2x+9+16=16$$

Add 9 and 16 to get 25.

$$2x^{2}+2x+25=16$$

Subtract 25 from both sides.

$$2x^{2}+2x=16-25$$

Subtract 25 from 16 to get -9.

$$2x^{2}+2x=-9$$

Divide both sides by 2.

$$\frac{2x^{2}+2x}{2}=-\frac{9}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{2}{2}x=-\frac{9}{2}$$

Divide 2 by 2.

$$x^{2}+x=-\frac{9}{2}$$

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

$$x^{2}+x+\left(\frac{1}{2}\right)^{2}=-\frac{9}{2}+\left(\frac{1}{2}\right)^{2}$$

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

$$x^{2}+x+\frac{1}{4}=-\frac{9}{2}+\frac{1}{4}$$

Add -\frac{9}{2} to \frac{1}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+x+\frac{1}{4}=-\frac{17}{4}$$

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

$$\left(x+\frac{1}{2}\right)^{2}=-\frac{17}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{-\frac{17}{4}}$$

Simplify.

$$x+\frac{1}{2}=\frac{\sqrt{17}i}{2} x+\frac{1}{2}=-\frac{\sqrt{17}i}{2}$$

Subtract \frac{1}{2} from both sides of the equation.

$$x=\frac{-1+\sqrt{17}i}{2} x=\frac{-\sqrt{17}i-1}{2}$$