Solve (2x^2+7x+4)

$$( 2 x ^ { 2 } + 7 x + 4 )$$

Factor

$$2\left(x-\frac{-\sqrt{17}-7}{4}\right)\left(x-\frac{\sqrt{17}-7}{4}\right)$$

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.

$$2x^{2}+7x+4=0$$

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.

$$x=\frac{-7±\sqrt{7^{2}-4\times 2\times 4}}{2\times 2}$$

Square 7.

$$x=\frac{-7±\sqrt{49-4\times 2\times 4}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-7±\sqrt{49-8\times 4}}{2\times 2}$$

Multiply -8 times 4.

$$x=\frac{-7±\sqrt{49-32}}{2\times 2}$$

Add 49 to -32.

$$x=\frac{-7±\sqrt{17}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-7±\sqrt{17}}{4}$$

Now solve the equation x=\frac{-7±\sqrt{17}}{4} when ± is plus. Add -7 to \sqrt{17}.

$$x=\frac{\sqrt{17}-7}{4}$$

Now solve the equation x=\frac{-7±\sqrt{17}}{4} when ± is minus. Subtract \sqrt{17} from -7.

$$x=\frac{-\sqrt{17}-7}{4}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-7+\sqrt{17}}{4} for x_{1} and \frac{-7-\sqrt{17}}{4} for x_{2}.

$$2x^{2}+7x+4=2\left(x-\frac{\sqrt{17}-7}{4}\right)\left(x-\frac{-\sqrt{17}-7}{4}\right)$$

Evaluate

$$2x^{2}+7x+4$$