Solve (-x-3)^2=(15-x^2)

$$( - x - 3 ) ^ { 2 } = ( 15 - x ^ { 2 } )$$

Solve for x

$$x=\frac{\sqrt{21}-3}{2}\approx 0.791287847$$

$$x=\frac{-\sqrt{21}-3}{2}\approx -3.791287847$$

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}.

$$\left(-x\right)^{2}-6\left(-x\right)+9=15-x^{2}$$

Calculate -x to the power of 2 and get x^{2}.

$$x^{2}-6\left(-x\right)+9=15-x^{2}$$

Multiply -6 and -1 to get 6.

$$x^{2}+6x+9=15-x^{2}$$

Subtract 15 from both sides.

$$x^{2}+6x+9-15=-x^{2}$$

Subtract 15 from 9 to get -6.

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

Add x^{2} to both sides.

$$x^{2}+6x-6+x^{2}=0$$

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

$$2x^{2}+6x-6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 6 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}-4\times 2\left(-6\right)}}{2\times 2}$$

Square 6.

$$x=\frac{-6±\sqrt{36-4\times 2\left(-6\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-6±\sqrt{36-8\left(-6\right)}}{2\times 2}$$

Multiply -8 times -6.

$$x=\frac{-6±\sqrt{36+48}}{2\times 2}$$

Add 36 to 48.

$$x=\frac{-6±\sqrt{84}}{2\times 2}$$

Take the square root of 84.

$$x=\frac{-6±2\sqrt{21}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-6±2\sqrt{21}}{4}$$

Now solve the equation x=\frac{-6±2\sqrt{21}}{4} when ± is plus. Add -6 to 2\sqrt{21}.

$$x=\frac{2\sqrt{21}-6}{4}$$

Divide -6+2\sqrt{21} by 4.

$$x=\frac{\sqrt{21}-3}{2}$$

Now solve the equation x=\frac{-6±2\sqrt{21}}{4} when ± is minus. Subtract 2\sqrt{21} from -6.

$$x=\frac{-2\sqrt{21}-6}{4}$$

Divide -6-2\sqrt{21} by 4.

$$x=\frac{-\sqrt{21}-3}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{21}-3}{2} x=\frac{-\sqrt{21}-3}{2}$$

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}.

$$\left(-x\right)^{2}-6\left(-x\right)+9=15-x^{2}$$

Calculate -x to the power of 2 and get x^{2}.

$$x^{2}-6\left(-x\right)+9=15-x^{2}$$

Multiply -6 and -1 to get 6.

$$x^{2}+6x+9=15-x^{2}$$

Add x^{2} to both sides.

$$x^{2}+6x+9+x^{2}=15$$

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

$$2x^{2}+6x+9=15$$

Subtract 9 from both sides.

$$2x^{2}+6x=15-9$$

Subtract 9 from 15 to get 6.

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

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

Divide 6 by 2.

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

Divide 6 by 2.

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

Divide 3, the coefficient of the x term, by 2 to get \frac{3}{2}. Then add the square of \frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square \frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+3x+\frac{9}{4}=3+\frac{9}{4}$$

Add 3 to \frac{9}{4}.

$$x^{2}+3x+\frac{9}{4}=\frac{21}{4}$$

Factor x^{2}+3x+\frac{9}{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}.

$$\left(x+\frac{3}{2}\right)^{2}=\frac{21}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{2}\right)^{2}}=\sqrt{\frac{21}{4}}$$

Simplify.

$$x+\frac{3}{2}=\frac{\sqrt{21}}{2} x+\frac{3}{2}=-\frac{\sqrt{21}}{2}$$

Subtract \frac{3}{2} from both sides of the equation.

$$x=\frac{\sqrt{21}-3}{2} x=\frac{-\sqrt{21}-3}{2}$$