Solve x^2+9x+3

$${ x }^{ 2 } +9x+3$$

Factor

$$\left(x-\frac{-\sqrt{69}-9}{2}\right)\left(x-\frac{\sqrt{69}-9}{2}\right)$$

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.

$$x^{2}+9x+3=0$$

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.

$$x=\frac{-9±\sqrt{9^{2}-4\times 3}}{2}$$

Square 9.

$$x=\frac{-9±\sqrt{81-4\times 3}}{2}$$

Multiply -4 times 3.

$$x=\frac{-9±\sqrt{81-12}}{2}$$

Add 81 to -12.

$$x=\frac{-9±\sqrt{69}}{2}$$

Now solve the equation x=\frac{-9±\sqrt{69}}{2} when ± is plus. Add -9 to \sqrt{69}.

$$x=\frac{\sqrt{69}-9}{2}$$

Now solve the equation x=\frac{-9±\sqrt{69}}{2} when ± is minus. Subtract \sqrt{69} from -9.

$$x=\frac{-\sqrt{69}-9}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-9+\sqrt{69}}{2} for x_{1} and \frac{-9-\sqrt{69}}{2} for x_{2}.

$$x^{2}+9x+3=\left(x-\frac{\sqrt{69}-9}{2}\right)\left(x-\frac{-\sqrt{69}-9}{2}\right)$$

Evaluate

$$x^{2}+9x+3$$