Solve x^2+11x-18

$${ x }^{ 2 } +11x-18$$

Factor

$$\left(x-\frac{-\sqrt{193}-11}{2}\right)\left(x-\frac{\sqrt{193}-11}{2}\right)$$

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.

$$x^{2}+11x-18=0$$

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.

$$x=\frac{-11±\sqrt{11^{2}-4\left(-18\right)}}{2}$$

Square 11.

$$x=\frac{-11±\sqrt{121-4\left(-18\right)}}{2}$$

Multiply -4 times -18.

$$x=\frac{-11±\sqrt{121+72}}{2}$$

Add 121 to 72.

$$x=\frac{-11±\sqrt{193}}{2}$$

Now solve the equation x=\frac{-11±\sqrt{193}}{2} when ± is plus. Add -11 to \sqrt{193}.

$$x=\frac{\sqrt{193}-11}{2}$$

Now solve the equation x=\frac{-11±\sqrt{193}}{2} when ± is minus. Subtract \sqrt{193} from -11.

$$x=\frac{-\sqrt{193}-11}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-11+\sqrt{193}}{2} for x_{1} and \frac{-11-\sqrt{193}}{2} for x_{2}.

$$x^{2}+11x-18=\left(x-\frac{\sqrt{193}-11}{2}\right)\left(x-\frac{-\sqrt{193}-11}{2}\right)$$

Evaluate

$$x^{2}+11x-18$$