Solve x^2+5x-9

$${ x }^{ 2 } +5x-9$$

Factor

$$\left(x-\frac{-\sqrt{61}-5}{2}\right)\left(x-\frac{\sqrt{61}-5}{2}\right)$$

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.

$$x^{2}+5x-9=0$$

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.

$$x=\frac{-5±\sqrt{5^{2}-4\left(-9\right)}}{2}$$

Square 5.

$$x=\frac{-5±\sqrt{25-4\left(-9\right)}}{2}$$

Multiply -4 times -9.

$$x=\frac{-5±\sqrt{25+36}}{2}$$

Add 25 to 36.

$$x=\frac{-5±\sqrt{61}}{2}$$

Now solve the equation x=\frac{-5±\sqrt{61}}{2} when ± is plus. Add -5 to \sqrt{61}.

$$x=\frac{\sqrt{61}-5}{2}$$

Now solve the equation x=\frac{-5±\sqrt{61}}{2} when ± is minus. Subtract \sqrt{61} from -5.

$$x=\frac{-\sqrt{61}-5}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-5+\sqrt{61}}{2} for x_{1} and \frac{-5-\sqrt{61}}{2} for x_{2}.

$$x^{2}+5x-9=\left(x-\frac{\sqrt{61}-5}{2}\right)\left(x-\frac{-\sqrt{61}-5}{2}\right)$$

Evaluate

$$x^{2}+5x-9$$