Solve 8-x-2x^2

$$8 - x - 2 x ^ { 2 }$$

Factor

$$-2\left(x-\frac{-\sqrt{65}-1}{4}\right)\left(x-\frac{\sqrt{65}-1}{4}\right)$$

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.

$$-2x^{2}-x+8=0$$

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.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\left(-2\right)\times 8}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-\left(-1\right)±\sqrt{1+8\times 8}}{2\left(-2\right)}$$

Multiply 8 times 8.

$$x=\frac{-\left(-1\right)±\sqrt{1+64}}{2\left(-2\right)}$$

Add 1 to 64.

$$x=\frac{-\left(-1\right)±\sqrt{65}}{2\left(-2\right)}$$

The opposite of -1 is 1.

$$x=\frac{1±\sqrt{65}}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{1±\sqrt{65}}{-4}$$

Now solve the equation x=\frac{1±\sqrt{65}}{-4} when ± is plus. Add 1 to \sqrt{65}.

$$x=\frac{\sqrt{65}+1}{-4}$$

Divide 1+\sqrt{65} by -4.

$$x=\frac{-\sqrt{65}-1}{4}$$

Now solve the equation x=\frac{1±\sqrt{65}}{-4} when ± is minus. Subtract \sqrt{65} from 1.

$$x=\frac{1-\sqrt{65}}{-4}$$

Divide 1-\sqrt{65} by -4.

$$x=\frac{\sqrt{65}-1}{4}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-1-\sqrt{65}}{4} for x_{1} and \frac{-1+\sqrt{65}}{4} for x_{2}.

$$-2x^{2}-x+8=-2\left(x-\frac{-\sqrt{65}-1}{4}\right)\left(x-\frac{\sqrt{65}-1}{4}\right)$$

Evaluate

$$8-x-2x^{2}$$