Solve x^2+40x+120

$${ x }^{ 2 } +40x+120$$

Factor

$$\left(x-\left(-2\sqrt{70}-20\right)\right)\left(x-\left(2\sqrt{70}-20\right)\right)$$

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.

$$x^{2}+40x+120=0$$

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.

$$x=\frac{-40±\sqrt{40^{2}-4\times 120}}{2}$$

Square 40.

$$x=\frac{-40±\sqrt{1600-4\times 120}}{2}$$

Multiply -4 times 120.

$$x=\frac{-40±\sqrt{1600-480}}{2}$$

Add 1600 to -480.

$$x=\frac{-40±\sqrt{1120}}{2}$$

Take the square root of 1120.

$$x=\frac{-40±4\sqrt{70}}{2}$$

Now solve the equation x=\frac{-40±4\sqrt{70}}{2} when ± is plus. Add -40 to 4\sqrt{70}.

$$x=\frac{4\sqrt{70}-40}{2}$$

Divide -40+4\sqrt{70} by 2.

$$x=2\sqrt{70}-20$$

Now solve the equation x=\frac{-40±4\sqrt{70}}{2} when ± is minus. Subtract 4\sqrt{70} from -40.

$$x=\frac{-4\sqrt{70}-40}{2}$$

Divide -40-4\sqrt{70} by 2.

$$x=-2\sqrt{70}-20$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -20+2\sqrt{70} for x_{1} and -20-2\sqrt{70} for x_{2}.

$$x^{2}+40x+120=\left(x-\left(2\sqrt{70}-20\right)\right)\left(x-\left(-2\sqrt{70}-20\right)\right)$$

Evaluate

$$x^{2}+40x+120$$