Solve 2x^2+2+9x

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

Factor

$$2\left(x-\frac{-\sqrt{65}-9}{4}\right)\left(x-\frac{\sqrt{65}-9}{4}\right)$$

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.

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

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.

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

Square 9.

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

Multiply -4 times 2.

$$x=\frac{-9±\sqrt{81-8\times 2}}{2\times 2}$$

Multiply -8 times 2.

$$x=\frac{-9±\sqrt{81-16}}{2\times 2}$$

Add 81 to -16.

$$x=\frac{-9±\sqrt{65}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-9±\sqrt{65}}{4}$$

Now solve the equation x=\frac{-9±\sqrt{65}}{4} when ± is plus. Add -9 to \sqrt{65}.

$$x=\frac{\sqrt{65}-9}{4}$$

Now solve the equation x=\frac{-9±\sqrt{65}}{4} when ± is minus. Subtract \sqrt{65} from -9.

$$x=\frac{-\sqrt{65}-9}{4}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-9+\sqrt{65}}{4} for x_{1} and \frac{-9-\sqrt{65}}{4} for x_{2}.

$$2x^{2}+9x+2=2\left(x-\frac{\sqrt{65}-9}{4}\right)\left(x-\frac{-\sqrt{65}-9}{4}\right)$$

Evaluate

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