Solve x^2-x-1=16180

$${ x }^{ 2 } -x-1=16180$$

Solve for x

$$x = \frac{5 \sqrt{2589} + 1}{2} \approx 127.705542332$$

$$x=\frac{1-5\sqrt{2589}}{2}\approx -126.705542332$$

Steps Using the Quadratic Formula

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.

$$x^{2}-x-1=16180$$

Subtract 16180 from both sides of the equation.

$$x^{2}-x-1-16180=16180-16180$$

Subtracting 16180 from itself leaves 0.

$$x^{2}-x-1-16180=0$$

Subtract 16180 from -1.

$$x^{2}-x-16181=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -1 for b, and -16181 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\left(-16181\right)}}{2}$$

Multiply -4 times -16181.

$$x=\frac{-\left(-1\right)±\sqrt{1+64724}}{2}$$

Add 1 to 64724.

$$x=\frac{-\left(-1\right)±\sqrt{64725}}{2}$$

Take the square root of 64725.

$$x=\frac{-\left(-1\right)±5\sqrt{2589}}{2}$$

The opposite of -1 is 1.

$$x=\frac{1±5\sqrt{2589}}{2}$$

Now solve the equation x=\frac{1±5\sqrt{2589}}{2} when ± is plus. Add 1 to 5\sqrt{2589}.

$$x=\frac{5\sqrt{2589}+1}{2}$$

Now solve the equation x=\frac{1±5\sqrt{2589}}{2} when ± is minus. Subtract 5\sqrt{2589} from 1.

$$x=\frac{1-5\sqrt{2589}}{2}$$

The equation is now solved.

$$x=\frac{5\sqrt{2589}+1}{2} x=\frac{1-5\sqrt{2589}}{2}$$

Steps for Completing the Square

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.

$$x^{2}-x-1=16180$$

Add 1 to both sides of the equation.

$$x^{2}-x-1-\left(-1\right)=16180-\left(-1\right)$$

Subtracting -1 from itself leaves 0.

$$x^{2}-x=16180-\left(-1\right)$$

Subtract -1 from 16180.

$$x^{2}-x=16181$$

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.

$$x^{2}-x+\left(-\frac{1}{2}\right)^{2}=16181+\left(-\frac{1}{2}\right)^{2}$$

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

$$x^{2}-x+\frac{1}{4}=16181+\frac{1}{4}$$

Add 16181 to \frac{1}{4}.

$$x^{2}-x+\frac{1}{4}=\frac{64725}{4}$$

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}.

$$\left(x-\frac{1}{2}\right)^{2}=\frac{64725}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{\frac{64725}{4}}$$

Simplify.

$$x-\frac{1}{2}=\frac{5\sqrt{2589}}{2} x-\frac{1}{2}=-\frac{5\sqrt{2589}}{2}$$

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

$$x=\frac{5\sqrt{2589}+1}{2} x=\frac{1-5\sqrt{2589}}{2}$$