Solve x^2-6x+4

$${ x }^{ 2 } -6x+4$$

Factor

$$\left(x-\left(3-\sqrt{5}\right)\right)\left(x-\left(\sqrt{5}+3\right)\right)$$

Steps Using the Quadratic Formula

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$x^{2}-6x+4=0$$

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.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 4}}{2}$$

Square -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\times 4}}{2}$$

Multiply -4 times 4.

$$x=\frac{-\left(-6\right)±\sqrt{36-16}}{2}$$

Add 36 to -16.

$$x=\frac{-\left(-6\right)±\sqrt{20}}{2}$$

Take the square root of 20.

$$x=\frac{-\left(-6\right)±2\sqrt{5}}{2}$$

The opposite of -6 is 6.

$$x=\frac{6±2\sqrt{5}}{2}$$

Now solve the equation x=\frac{6±2\sqrt{5}}{2} when ± is plus. Add 6 to 2\sqrt{5}.

$$x=\frac{2\sqrt{5}+6}{2}$$

Divide 6+2\sqrt{5} by 2.

$$x=\sqrt{5}+3$$

Now solve the equation x=\frac{6±2\sqrt{5}}{2} when ± is minus. Subtract 2\sqrt{5} from 6.

$$x=\frac{6-2\sqrt{5}}{2}$$

Divide 6-2\sqrt{5} by 2.

$$x=3-\sqrt{5}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 3+\sqrt{5} for x_{1} and 3-\sqrt{5} for x_{2}.

$$x^{2}-6x+4=\left(x-\left(\sqrt{5}+3\right)\right)\left(x-\left(3-\sqrt{5}\right)\right)$$

Evaluate

$$x^{2}-6x+4$$