Solve 2^2+x^2=x

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

Solve for x (complex solution)

$$x=\frac{1+\sqrt{15}i}{2}\approx 0.5+1.936491673i$$

$$x=\frac{-\sqrt{15}i+1}{2}\approx 0.5-1.936491673i$$

Steps Using the Quadratic Formula

Calculate 2 to the power of 2 and get 4.

$$4+x^{2}=x$$

Subtract x from both sides.

$$4+x^{2}-x=0$$

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.

$$x^{2}-x+4=0$$

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

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

Multiply -4 times 4.

$$x=\frac{-\left(-1\right)±\sqrt{1-16}}{2}$$

Add 1 to -16.

$$x=\frac{-\left(-1\right)±\sqrt{-15}}{2}$$

Take the square root of -15.

$$x=\frac{-\left(-1\right)±\sqrt{15}i}{2}$$

The opposite of -1 is 1.

$$x=\frac{1±\sqrt{15}i}{2}$$

Now solve the equation x=\frac{1±\sqrt{15}i}{2} when ± is plus. Add 1 to i\sqrt{15}.

$$x=\frac{1+\sqrt{15}i}{2}$$

Now solve the equation x=\frac{1±\sqrt{15}i}{2} when ± is minus. Subtract i\sqrt{15} from 1.

$$x=\frac{-\sqrt{15}i+1}{2}$$

The equation is now solved.

$$x=\frac{1+\sqrt{15}i}{2} x=\frac{-\sqrt{15}i+1}{2}$$

Steps for Completing the Square

Calculate 2 to the power of 2 and get 4.

$$4+x^{2}=x$$

Subtract x from both sides.

$$4+x^{2}-x=0$$

Subtract 4 from both sides. Anything subtracted from zero gives its negation.

$$x^{2}-x=-4$$

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.

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

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

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

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

$$x^{2}-x+\frac{1}{4}=-\frac{15}{4}$$

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}.

$$\left(x-\frac{1}{2}\right)^{2}=-\frac{15}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{-\frac{15}{4}}$$

Simplify.

$$x-\frac{1}{2}=\frac{\sqrt{15}i}{2} x-\frac{1}{2}=-\frac{\sqrt{15}i}{2}$$

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

$$x=\frac{1+\sqrt{15}i}{2} x=\frac{-\sqrt{15}i+1}{2}$$