Solve x^2+7=4x

$$x ^ { 2 } + 7 = 4 x$$

Solve for x (complex solution)

$$x=2+\sqrt{3}i\approx 2+1.732050808i$$

$$x=-\sqrt{3}i+2\approx 2-1.732050808i$$

Steps Using the Quadratic Formula

Subtract 4x from both sides.

$$x^{2}+7-4x=0$$

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.

$$x^{2}-4x+7=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -4 for b, and 7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 7}}{2}$$

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\times 7}}{2}$$

Multiply -4 times 7.

$$x=\frac{-\left(-4\right)±\sqrt{16-28}}{2}$$

Add 16 to -28.

$$x=\frac{-\left(-4\right)±\sqrt{-12}}{2}$$

Take the square root of -12.

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

The opposite of -4 is 4.

$$x=\frac{4±2\sqrt{3}i}{2}$$

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

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

Divide 4+2i\sqrt{3} by 2.

$$x=2+\sqrt{3}i$$

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

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

Divide 4-2i\sqrt{3} by 2.

$$x=-\sqrt{3}i+2$$

The equation is now solved.

$$x=2+\sqrt{3}i x=-\sqrt{3}i+2$$

Steps for Completing the Square

Subtract 4x from both sides.

$$x^{2}+7-4x=0$$

Subtract 7 from both sides. Anything subtracted from zero gives its negation.

$$x^{2}-4x=-7$$

Divide -4, the coefficient of the x term, by 2 to get -2. Then add the square of -2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-4x+\left(-2\right)^{2}=-7+\left(-2\right)^{2}$$

Square -2.

$$x^{2}-4x+4=-7+4$$

Add -7 to 4.

$$x^{2}-4x+4=-3$$

Factor x^{2}-4x+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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-2=\sqrt{3}i x-2=-\sqrt{3}i$$

Add 2 to both sides of the equation.

$$x=2+\sqrt{3}i x=-\sqrt{3}i+2$$