Solve x^2-4x-5*8

$$x ^ { 2 } - 4 x - 5 \cdot 8$$

Factor

$$\left(x-\left(2-2\sqrt{11}\right)\right)\left(x-\left(2\sqrt{11}+2\right)\right)$$

Steps Using the Quadratic Formula

Multiply 5 and 8 to get 40.

$$factor(x^{2}-4x-40)$$

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.

$$x^{2}-4x-40=0$$

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.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\left(-40\right)}}{2}$$

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\left(-40\right)}}{2}$$

Multiply -4 times -40.

$$x=\frac{-\left(-4\right)±\sqrt{16+160}}{2}$$

Add 16 to 160.

$$x=\frac{-\left(-4\right)±\sqrt{176}}{2}$$

Take the square root of 176.

$$x=\frac{-\left(-4\right)±4\sqrt{11}}{2}$$

The opposite of -4 is 4.

$$x=\frac{4±4\sqrt{11}}{2}$$

Now solve the equation x=\frac{4±4\sqrt{11}}{2} when ± is plus. Add 4 to 4\sqrt{11}.

$$x=\frac{4\sqrt{11}+4}{2}$$

Divide 4+4\sqrt{11} by 2.

$$x=2\sqrt{11}+2$$

Now solve the equation x=\frac{4±4\sqrt{11}}{2} when ± is minus. Subtract 4\sqrt{11} from 4.

$$x=\frac{4-4\sqrt{11}}{2}$$

Divide 4-4\sqrt{11} by 2.

$$x=2-2\sqrt{11}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 2+2\sqrt{11} for x_{1} and 2-2\sqrt{11} for x_{2}.

$$x^{2}-4x-40=\left(x-\left(2\sqrt{11}+2\right)\right)\left(x-\left(2-2\sqrt{11}\right)\right)$$

Evaluate

$$x^{2}-4x-40$$

Solution Steps

Multiply 5 and 8 to get 40.

$$x^{2}-4x-40$$