Solve x^2+2-5x

$$x ^ { 2 } + 2 - 5 x$$

Factor

$$\left(x-\frac{5-\sqrt{17}}{2}\right)\left(x-\frac{\sqrt{17}+5}{2}\right)$$

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.

$$x^{2}-5x+2=0$$

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.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 2}}{2}$$

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 2}}{2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-5\right)±\sqrt{25-8}}{2}$$

Add 25 to -8.

$$x=\frac{-\left(-5\right)±\sqrt{17}}{2}$$

The opposite of -5 is 5.

$$x=\frac{5±\sqrt{17}}{2}$$

Now solve the equation x=\frac{5±\sqrt{17}}{2} when ± is plus. Add 5 to \sqrt{17}.

$$x=\frac{\sqrt{17}+5}{2}$$

Now solve the equation x=\frac{5±\sqrt{17}}{2} when ± is minus. Subtract \sqrt{17} from 5.

$$x=\frac{5-\sqrt{17}}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{5+\sqrt{17}}{2} for x_{1} and \frac{5-\sqrt{17}}{2} for x_{2}.

$$x^{2}-5x+2=\left(x-\frac{\sqrt{17}+5}{2}\right)\left(x-\frac{5-\sqrt{17}}{2}\right)$$

Evaluate

$$x^{2}-5x+2$$