Solve x^2+2x+1-5*4

$${ x }^{ 2 } +2x+1-5 \times 4$$

Evaluate

$$x^{2}+2x-19$$

Solution Steps

Multiply 5 and 4 to get 20.

$$x^{2}+2x+1-20$$

Subtract 20 from 1 to get -19.

$$x^{2}+2x-19$$

Factor

$$\left(x-\left(-2\sqrt{5}-1\right)\right)\left(x-\left(2\sqrt{5}-1\right)\right)$$

Steps Using the Quadratic Formula

Multiply 5 and 4 to get 20.

$$factor(x^{2}+2x+1-20)$$

Subtract 20 from 1 to get -19.

$$factor(x^{2}+2x-19)$$

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.

$$x^{2}+2x-19=0$$

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.

$$x=\frac{-2±\sqrt{2^{2}-4\left(-19\right)}}{2}$$

Square 2.

$$x=\frac{-2±\sqrt{4-4\left(-19\right)}}{2}$$

Multiply -4 times -19.

$$x=\frac{-2±\sqrt{4+76}}{2}$$

Add 4 to 76.

$$x=\frac{-2±\sqrt{80}}{2}$$

Take the square root of 80.

$$x=\frac{-2±4\sqrt{5}}{2}$$

Now solve the equation x=\frac{-2±4\sqrt{5}}{2} when ± is plus. Add -2 to 4\sqrt{5}.

$$x=\frac{4\sqrt{5}-2}{2}$$

Divide -2+4\sqrt{5} by 2.

$$x=2\sqrt{5}-1$$

Now solve the equation x=\frac{-2±4\sqrt{5}}{2} when ± is minus. Subtract 4\sqrt{5} from -2.

$$x=\frac{-4\sqrt{5}-2}{2}$$

Divide -2-4\sqrt{5} by 2.

$$x=-2\sqrt{5}-1$$

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

$$x^{2}+2x-19=\left(x-\left(2\sqrt{5}-1\right)\right)\left(x-\left(-2\sqrt{5}-1\right)\right)$$