Solve x^2+19x+14

$${ x }^{ 2 } +19x+14$$

Factor

$$\left(x-\frac{-\sqrt{305}-19}{2}\right)\left(x-\frac{\sqrt{305}-19}{2}\right)$$

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.

$$x^{2}+19x+14=0$$

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.

$$x=\frac{-19±\sqrt{19^{2}-4\times 14}}{2}$$

Square 19.

$$x=\frac{-19±\sqrt{361-4\times 14}}{2}$$

Multiply -4 times 14.

$$x=\frac{-19±\sqrt{361-56}}{2}$$

Add 361 to -56.

$$x=\frac{-19±\sqrt{305}}{2}$$

Now solve the equation x=\frac{-19±\sqrt{305}}{2} when ± is plus. Add -19 to \sqrt{305}.

$$x=\frac{\sqrt{305}-19}{2}$$

Now solve the equation x=\frac{-19±\sqrt{305}}{2} when ± is minus. Subtract \sqrt{305} from -19.

$$x=\frac{-\sqrt{305}-19}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-19+\sqrt{305}}{2} for x_{1} and \frac{-19-\sqrt{305}}{2} for x_{2}.

$$x^{2}+19x+14=\left(x-\frac{\sqrt{305}-19}{2}\right)\left(x-\frac{-\sqrt{305}-19}{2}\right)$$

Evaluate

$$x^{2}+19x+14$$