Solve 20x+2x^2+10

$$20x+2 { x }^{ 2 } +10$$

Factor

$$2\left(x-\left(-2\sqrt{5}-5\right)\right)\left(x-\left(2\sqrt{5}-5\right)\right)$$

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.

$$2x^{2}+20x+10=0$$

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.

$$x=\frac{-20±\sqrt{20^{2}-4\times 2\times 10}}{2\times 2}$$

Square 20.

$$x=\frac{-20±\sqrt{400-4\times 2\times 10}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-20±\sqrt{400-8\times 10}}{2\times 2}$$

Multiply -8 times 10.

$$x=\frac{-20±\sqrt{400-80}}{2\times 2}$$

Add 400 to -80.

$$x=\frac{-20±\sqrt{320}}{2\times 2}$$

Take the square root of 320.

$$x=\frac{-20±8\sqrt{5}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-20±8\sqrt{5}}{4}$$

Now solve the equation x=\frac{-20±8\sqrt{5}}{4} when ± is plus. Add -20 to 8\sqrt{5}.

$$x=\frac{8\sqrt{5}-20}{4}$$

Divide -20+8\sqrt{5} by 4.

$$x=2\sqrt{5}-5$$

Now solve the equation x=\frac{-20±8\sqrt{5}}{4} when ± is minus. Subtract 8\sqrt{5} from -20.

$$x=\frac{-8\sqrt{5}-20}{4}$$

Divide -20-8\sqrt{5} by 4.

$$x=-2\sqrt{5}-5$$

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

$$2x^{2}+20x+10=2\left(x-\left(2\sqrt{5}-5\right)\right)\left(x-\left(-2\sqrt{5}-5\right)\right)$$

Evaluate

$$2\left(x^{2}+10x+5\right)$$