Solve 2x^2-x-5

$$2 { x }^{ 2 } -x-5$$

Factor

$$2\left(x-\frac{1-\sqrt{41}}{4}\right)\left(x-\frac{\sqrt{41}+1}{4}\right)$$

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.

$$2x^{2}-x-5=0$$

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.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\times 2\left(-5\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-1\right)±\sqrt{1-8\left(-5\right)}}{2\times 2}$$

Multiply -8 times -5.

$$x=\frac{-\left(-1\right)±\sqrt{1+40}}{2\times 2}$$

Add 1 to 40.

$$x=\frac{-\left(-1\right)±\sqrt{41}}{2\times 2}$$

The opposite of -1 is 1.

$$x=\frac{1±\sqrt{41}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{1±\sqrt{41}}{4}$$

Now solve the equation x=\frac{1±\sqrt{41}}{4} when ± is plus. Add 1 to \sqrt{41}.

$$x=\frac{\sqrt{41}+1}{4}$$

Now solve the equation x=\frac{1±\sqrt{41}}{4} when ± is minus. Subtract \sqrt{41} from 1.

$$x=\frac{1-\sqrt{41}}{4}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{1+\sqrt{41}}{4} for x_{1} and \frac{1-\sqrt{41}}{4} for x_{2}.

$$2x^{2}-x-5=2\left(x-\frac{\sqrt{41}+1}{4}\right)\left(x-\frac{1-\sqrt{41}}{4}\right)$$

Evaluate

$$2x^{2}-x-5$$