Solve 5x^2+x-2

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

Factor

$$5\left(x-\frac{-\sqrt{41}-1}{10}\right)\left(x-\frac{\sqrt{41}-1}{10}\right)$$

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.

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

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.

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

Square 1.

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

Multiply -4 times 5.

$$x=\frac{-1±\sqrt{1-20\left(-2\right)}}{2\times 5}$$

Multiply -20 times -2.

$$x=\frac{-1±\sqrt{1+40}}{2\times 5}$$

Add 1 to 40.

$$x=\frac{-1±\sqrt{41}}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{-1±\sqrt{41}}{10}$$

Now solve the equation x=\frac{-1±\sqrt{41}}{10} when ± is plus. Add -1 to \sqrt{41}.

$$x=\frac{\sqrt{41}-1}{10}$$

Now solve the equation x=\frac{-1±\sqrt{41}}{10} when ± is minus. Subtract \sqrt{41} from -1.

$$x=\frac{-\sqrt{41}-1}{10}$$

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}}{10} for x_{1} and \frac{-1-\sqrt{41}}{10} for x_{2}.

$$5x^{2}+x-2=5\left(x-\frac{\sqrt{41}-1}{10}\right)\left(x-\frac{-\sqrt{41}-1}{10}\right)$$

Evaluate

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