Solve x^2-42x+45=

$${ x }^{ 2 } -42x+45=$$

Factor

$$\left(x-\left(21-6\sqrt{11}\right)\right)\left(x-\left(6\sqrt{11}+21\right)\right)$$

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.

$$x^{2}-42x+45=0$$

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.

$$x=\frac{-\left(-42\right)±\sqrt{\left(-42\right)^{2}-4\times 45}}{2}$$

Square -42.

$$x=\frac{-\left(-42\right)±\sqrt{1764-4\times 45}}{2}$$

Multiply -4 times 45.

$$x=\frac{-\left(-42\right)±\sqrt{1764-180}}{2}$$

Add 1764 to -180.

$$x=\frac{-\left(-42\right)±\sqrt{1584}}{2}$$

Take the square root of 1584.

$$x=\frac{-\left(-42\right)±12\sqrt{11}}{2}$$

The opposite of -42 is 42.

$$x=\frac{42±12\sqrt{11}}{2}$$

Now solve the equation x=\frac{42±12\sqrt{11}}{2} when ± is plus. Add 42 to 12\sqrt{11}.

$$x=\frac{12\sqrt{11}+42}{2}$$

Divide 42+12\sqrt{11} by 2.

$$x=6\sqrt{11}+21$$

Now solve the equation x=\frac{42±12\sqrt{11}}{2} when ± is minus. Subtract 12\sqrt{11} from 42.

$$x=\frac{42-12\sqrt{11}}{2}$$

Divide 42-12\sqrt{11} by 2.

$$x=21-6\sqrt{11}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 21+6\sqrt{11} for x_{1} and 21-6\sqrt{11} for x_{2}.

$$x^{2}-42x+45=\left(x-\left(6\sqrt{11}+21\right)\right)\left(x-\left(21-6\sqrt{11}\right)\right)$$

Evaluate

$$x^{2}-42x+45$$