Solve x^2-12x-18

$${ x }^{ 2 } -12x-18$$

Factor

$$\left(x-\left(6-3\sqrt{6}\right)\right)\left(x-\left(3\sqrt{6}+6\right)\right)$$

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.

$$x^{2}-12x-18=0$$

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.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\left(-18\right)}}{2}$$

Square -12.

$$x=\frac{-\left(-12\right)±\sqrt{144-4\left(-18\right)}}{2}$$

Multiply -4 times -18.

$$x=\frac{-\left(-12\right)±\sqrt{144+72}}{2}$$

Add 144 to 72.

$$x=\frac{-\left(-12\right)±\sqrt{216}}{2}$$

Take the square root of 216.

$$x=\frac{-\left(-12\right)±6\sqrt{6}}{2}$$

The opposite of -12 is 12.

$$x=\frac{12±6\sqrt{6}}{2}$$

Now solve the equation x=\frac{12±6\sqrt{6}}{2} when ± is plus. Add 12 to 6\sqrt{6}.

$$x=\frac{6\sqrt{6}+12}{2}$$

Divide 12+6\sqrt{6} by 2.

$$x=3\sqrt{6}+6$$

Now solve the equation x=\frac{12±6\sqrt{6}}{2} when ± is minus. Subtract 6\sqrt{6} from 12.

$$x=\frac{12-6\sqrt{6}}{2}$$

Divide 12-6\sqrt{6} by 2.

$$x=6-3\sqrt{6}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 6+3\sqrt{6} for x_{1} and 6-3\sqrt{6} for x_{2}.

$$x^{2}-12x-18=\left(x-\left(3\sqrt{6}+6\right)\right)\left(x-\left(6-3\sqrt{6}\right)\right)$$

Evaluate

$$x^{2}-12x-18$$