Solve (x-2)=x^2+(sqrt{3}-1)

$$( x - 2 ) = x ^ { 2 } + ( \sqrt { 3 } - 1 )$$

Solve for x (complex solution)

$$x=\frac{-i\sqrt{4\sqrt{3}+3}+1}{2}\approx 0.5-1.575452572i$$

$$x=\frac{1+i\sqrt{4\sqrt{3}+3}}{2}\approx 0.5+1.575452572i$$

Steps Using the Quadratic Formula

Subtract x^{2} from both sides.

$$x-2-x^{2}=\sqrt{3}-1$$

Subtract \sqrt{3} from both sides.

$$x-2-x^{2}-\sqrt{3}=-1$$

Add 1 to both sides.

$$x-2-x^{2}-\sqrt{3}+1=0$$

Add -2 and 1 to get -1.

$$x-1-x^{2}-\sqrt{3}=0$$

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.

$$-x^{2}+x-\sqrt{3}-1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 1 for b, and -1-\sqrt{3} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-1±\sqrt{1^{2}-4\left(-1\right)\left(-\sqrt{3}-1\right)}}{2\left(-1\right)}$$

Square 1.

$$x=\frac{-1±\sqrt{1-4\left(-1\right)\left(-\sqrt{3}-1\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-1±\sqrt{1+4\left(-\sqrt{3}-1\right)}}{2\left(-1\right)}$$

Multiply 4 times -1-\sqrt{3}.

$$x=\frac{-1±\sqrt{1-4\sqrt{3}-4}}{2\left(-1\right)}$$

Add 1 to -4-4\sqrt{3}.

$$x=\frac{-1±\sqrt{-4\sqrt{3}-3}}{2\left(-1\right)}$$

Take the square root of -3-4\sqrt{3}.

$$x=\frac{-1±i\sqrt{4\sqrt{3}+3}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-1±i\sqrt{4\sqrt{3}+3}}{-2}$$

Now solve the equation x=\frac{-1±i\sqrt{4\sqrt{3}+3}}{-2} when ± is plus. Add -1 to i\sqrt{3+4\sqrt{3}}.

$$x=\frac{\sqrt[4]{3}i\sqrt{\sqrt{3}+4}-1}{-2}$$

Divide -1+i\sqrt[4]{3}\sqrt{\sqrt{3}+4} by -2.

$$x=\frac{-\sqrt[4]{3}i\sqrt{\sqrt{3}+4}+1}{2}$$

Now solve the equation x=\frac{-1±i\sqrt{4\sqrt{3}+3}}{-2} when ± is minus. Subtract i\sqrt{3+4\sqrt{3}} from -1.

$$x=\frac{-\sqrt[4]{3}i\sqrt{\sqrt{3}+4}-1}{-2}$$

Divide -1-i\sqrt[4]{3}\sqrt{\sqrt{3}+4} by -2.

$$x=\frac{\sqrt[4]{3}i\sqrt{\sqrt{3}+4}+1}{2}$$

The equation is now solved.

$$x=\frac{-\sqrt[4]{3}i\sqrt{\sqrt{3}+4}+1}{2} x=\frac{\sqrt[4]{3}i\sqrt{\sqrt{3}+4}+1}{2}$$

Steps for Completing the Square

Subtract x^{2} from both sides.

$$x-2-x^{2}=\sqrt{3}-1$$

Add 2 to both sides.

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

Add -1 and 2 to get 1.

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

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.

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

Divide both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

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

Divide 1 by -1.

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

Divide \sqrt{3}+1 by -1.

$$x^{2}-x=-\left(\sqrt{3}+1\right)$$

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.

$$x^{2}-x+\left(-\frac{1}{2}\right)^{2}=-\left(\sqrt{3}+1\right)+\left(-\frac{1}{2}\right)^{2}$$

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

$$x^{2}-x+\frac{1}{4}=-\left(\sqrt{3}+1\right)+\frac{1}{4}$$

Add -\left(\sqrt{3}+1\right) to \frac{1}{4}.

$$x^{2}-x+\frac{1}{4}=-\sqrt{3}-\frac{3}{4}$$

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}.

$$\left(x-\frac{1}{2}\right)^{2}=-\sqrt{3}-\frac{3}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{-\sqrt{3}-\frac{3}{4}}$$

Simplify.

$$x-\frac{1}{2}=\frac{i\sqrt{4\sqrt{3}+3}}{2} x-\frac{1}{2}=-\frac{\sqrt[4]{3}i\sqrt{\sqrt{3}+4}}{2}$$

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

$$x=\frac{\sqrt[4]{3}i\sqrt{\sqrt{3}+4}+1}{2} x=\frac{-\sqrt[4]{3}i\sqrt{\sqrt{3}+4}+1}{2}$$