Solve x^2-2x-15=-x^2+16

$$x ^ { 2 } - 2 x - 15 = - x ^ { 2 } + 16$$

Solve for x

$$x = \frac{3 \sqrt{7} + 1}{2} \approx 4.468626967$$

$$x=\frac{1-3\sqrt{7}}{2}\approx -3.468626967$$

Steps Using the Quadratic Formula

Add x^{2} to both sides.

$$x^{2}-2x-15+x^{2}=16$$

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

$$2x^{2}-2x-15=16$$

Subtract 16 from both sides.

$$2x^{2}-2x-15-16=0$$

Subtract 16 from -15 to get -31.

$$2x^{2}-2x-31=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -2 for b, and -31 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 2\left(-31\right)}}{2\times 2}$$

Square -2.

$$x=\frac{-\left(-2\right)±\sqrt{4-4\times 2\left(-31\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-2\right)±\sqrt{4-8\left(-31\right)}}{2\times 2}$$

Multiply -8 times -31.

$$x=\frac{-\left(-2\right)±\sqrt{4+248}}{2\times 2}$$

Add 4 to 248.

$$x=\frac{-\left(-2\right)±\sqrt{252}}{2\times 2}$$

Take the square root of 252.

$$x=\frac{-\left(-2\right)±6\sqrt{7}}{2\times 2}$$

The opposite of -2 is 2.

$$x=\frac{2±6\sqrt{7}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{2±6\sqrt{7}}{4}$$

Now solve the equation x=\frac{2±6\sqrt{7}}{4} when ± is plus. Add 2 to 6\sqrt{7}.

$$x=\frac{6\sqrt{7}+2}{4}$$

Divide 2+6\sqrt{7} by 4.

$$x=\frac{3\sqrt{7}+1}{2}$$

Now solve the equation x=\frac{2±6\sqrt{7}}{4} when ± is minus. Subtract 6\sqrt{7} from 2.

$$x=\frac{2-6\sqrt{7}}{4}$$

Divide 2-6\sqrt{7} by 4.

$$x=\frac{1-3\sqrt{7}}{2}$$

The equation is now solved.

$$x=\frac{3\sqrt{7}+1}{2} x=\frac{1-3\sqrt{7}}{2}$$

Steps for Completing the Square

Add x^{2} to both sides.

$$x^{2}-2x-15+x^{2}=16$$

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

$$2x^{2}-2x-15=16$$

Add 15 to both sides.

$$2x^{2}-2x=16+15$$

Add 16 and 15 to get 31.

$$2x^{2}-2x=31$$

Divide both sides by 2.

$$\frac{2x^{2}-2x}{2}=\frac{31}{2}$$

Dividing by 2 undoes the multiplication by 2.

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

Divide -2 by 2.

$$x^{2}-x=\frac{31}{2}$$

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.

$$x^{2}-x+\left(-\frac{1}{2}\right)^{2}=\frac{31}{2}+\left(-\frac{1}{2}\right)^{2}$$

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

$$x^{2}-x+\frac{1}{4}=\frac{31}{2}+\frac{1}{4}$$

Add \frac{31}{2} to \frac{1}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-x+\frac{1}{4}=\frac{63}{4}$$

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}.

$$\left(x-\frac{1}{2}\right)^{2}=\frac{63}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{\frac{63}{4}}$$

Simplify.

$$x-\frac{1}{2}=\frac{3\sqrt{7}}{2} x-\frac{1}{2}=-\frac{3\sqrt{7}}{2}$$

Add \frac{1}{2} to both sides of the equation.

$$x=\frac{3\sqrt{7}+1}{2} x=\frac{1-3\sqrt{7}}{2}$$