Solve 0=x^2-4x+29

$$0 = x ^ { 2 } - 4 x + 29$$

Solve for x (complex solution)

$$x=2+5i$$

$$x=2-5i$$

Steps Using the Quadratic Formula

Swap sides so that all variable terms are on the left hand side.

$$x^{2}-4x+29=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -4 for b, and 29 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 29}}{2}$$

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\times 29}}{2}$$

Multiply -4 times 29.

$$x=\frac{-\left(-4\right)±\sqrt{16-116}}{2}$$

Add 16 to -116.

$$x=\frac{-\left(-4\right)±\sqrt{-100}}{2}$$

Take the square root of -100.

$$x=\frac{-\left(-4\right)±10i}{2}$$

The opposite of -4 is 4.

$$x=\frac{4±10i}{2}$$

Now solve the equation x=\frac{4±10i}{2} when ± is plus. Add 4 to 10i.

$$x=\frac{4+10i}{2}$$

Divide 4+10i by 2.

$$x=2+5i$$

Now solve the equation x=\frac{4±10i}{2} when ± is minus. Subtract 10i from 4.

$$x=\frac{4-10i}{2}$$

Divide 4-10i by 2.

$$x=2-5i$$

The equation is now solved.

$$x=2+5i x=2-5i$$

Steps for Completing the Square

Swap sides so that all variable terms are on the left hand side.

$$x^{2}-4x+29=0$$

Subtract 29 from both sides. Anything subtracted from zero gives its negation.

$$x^{2}-4x=-29$$

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.

$$x^{2}-4x+\left(-2\right)^{2}=-29+\left(-2\right)^{2}$$

Square -2.

$$x^{2}-4x+4=-29+4$$

Add -29 to 4.

$$x^{2}-4x+4=-25$$

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}.

$$\left(x-2\right)^{2}=-25$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-2\right)^{2}}=\sqrt{-25}$$

Simplify.

$$x-2=5i x-2=-5i$$

Add 2 to both sides of the equation.

$$x=2+5i x=2-5i$$