Solve x^2-3x+1=

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

Factor

$$\left(x-\frac{3-\sqrt{5}}{2}\right)\left(x-\frac{\sqrt{5}+3}{2}\right)$$

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.

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

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.

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

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4}}{2}$$

Add 9 to -4.

$$x=\frac{-\left(-3\right)±\sqrt{5}}{2}$$

The opposite of -3 is 3.

$$x=\frac{3±\sqrt{5}}{2}$$

Now solve the equation x=\frac{3±\sqrt{5}}{2} when ± is plus. Add 3 to \sqrt{5}.

$$x=\frac{\sqrt{5}+3}{2}$$

Now solve the equation x=\frac{3±\sqrt{5}}{2} when ± is minus. Subtract \sqrt{5} from 3.

$$x=\frac{3-\sqrt{5}}{2}$$

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

$$x^{2}-3x+1=\left(x-\frac{\sqrt{5}+3}{2}\right)\left(x-\frac{3-\sqrt{5}}{2}\right)$$

Evaluate

$$x^{2}-3x+1$$