Solve x^2+12x-9

$${ x }^{ 2 } +12x-9$$

Factor

$$\left(x-\left(-3\sqrt{5}-6\right)\right)\left(x-\left(3\sqrt{5}-6\right)\right)$$

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.

$$x^{2}+12x-9=0$$

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.

$$x=\frac{-12±\sqrt{12^{2}-4\left(-9\right)}}{2}$$

Square 12.

$$x=\frac{-12±\sqrt{144-4\left(-9\right)}}{2}$$

Multiply -4 times -9.

$$x=\frac{-12±\sqrt{144+36}}{2}$$

Add 144 to 36.

$$x=\frac{-12±\sqrt{180}}{2}$$

Take the square root of 180.

$$x=\frac{-12±6\sqrt{5}}{2}$$

Now solve the equation x=\frac{-12±6\sqrt{5}}{2} when ± is plus. Add -12 to 6\sqrt{5}.

$$x=\frac{6\sqrt{5}-12}{2}$$

Divide -12+6\sqrt{5} by 2.

$$x=3\sqrt{5}-6$$

Now solve the equation x=\frac{-12±6\sqrt{5}}{2} when ± is minus. Subtract 6\sqrt{5} from -12.

$$x=\frac{-6\sqrt{5}-12}{2}$$

Divide -12-6\sqrt{5} by 2.

$$x=-3\sqrt{5}-6$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -6+3\sqrt{5} for x_{1} and -6-3\sqrt{5} for x_{2}.

$$x^{2}+12x-9=\left(x-\left(3\sqrt{5}-6\right)\right)\left(x-\left(-3\sqrt{5}-6\right)\right)$$

Evaluate

$$x^{2}+12x-9$$