Solve 2x^2+10x+5

$$2 { x }^{ 2 } +10x+5$$

Factor

$$2\left(x-\frac{-\sqrt{15}-5}{2}\right)\left(x-\frac{\sqrt{15}-5}{2}\right)$$

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.

$$2x^{2}+10x+5=0$$

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.

$$x=\frac{-10±\sqrt{10^{2}-4\times 2\times 5}}{2\times 2}$$

Square 10.

$$x=\frac{-10±\sqrt{100-4\times 2\times 5}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-10±\sqrt{100-8\times 5}}{2\times 2}$$

Multiply -8 times 5.

$$x=\frac{-10±\sqrt{100-40}}{2\times 2}$$

Add 100 to -40.

$$x=\frac{-10±\sqrt{60}}{2\times 2}$$

Take the square root of 60.

$$x=\frac{-10±2\sqrt{15}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-10±2\sqrt{15}}{4}$$

Now solve the equation x=\frac{-10±2\sqrt{15}}{4} when ± is plus. Add -10 to 2\sqrt{15}.

$$x=\frac{2\sqrt{15}-10}{4}$$

Divide -10+2\sqrt{15} by 4.

$$x=\frac{\sqrt{15}-5}{2}$$

Now solve the equation x=\frac{-10±2\sqrt{15}}{4} when ± is minus. Subtract 2\sqrt{15} from -10.

$$x=\frac{-2\sqrt{15}-10}{4}$$

Divide -10-2\sqrt{15} by 4.

$$x=\frac{-\sqrt{15}-5}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-5+\sqrt{15}}{2} for x_{1} and \frac{-5-\sqrt{15}}{2} for x_{2}.

$$2x^{2}+10x+5=2\left(x-\frac{\sqrt{15}-5}{2}\right)\left(x-\frac{-\sqrt{15}-5}{2}\right)$$

Evaluate

$$2x^{2}+10x+5$$