Solve 5x^2+2x-6

$$5 { x }^{ 2 } +2x-6$$

Factor

$$5\left(x-\frac{-\sqrt{31}-1}{5}\right)\left(x-\frac{\sqrt{31}-1}{5}\right)$$

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.

$$5x^{2}+2x-6=0$$

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.

$$x=\frac{-2±\sqrt{2^{2}-4\times 5\left(-6\right)}}{2\times 5}$$

Square 2.

$$x=\frac{-2±\sqrt{4-4\times 5\left(-6\right)}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-2±\sqrt{4-20\left(-6\right)}}{2\times 5}$$

Multiply -20 times -6.

$$x=\frac{-2±\sqrt{4+120}}{2\times 5}$$

Add 4 to 120.

$$x=\frac{-2±\sqrt{124}}{2\times 5}$$

Take the square root of 124.

$$x=\frac{-2±2\sqrt{31}}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{-2±2\sqrt{31}}{10}$$

Now solve the equation x=\frac{-2±2\sqrt{31}}{10} when ± is plus. Add -2 to 2\sqrt{31}.

$$x=\frac{2\sqrt{31}-2}{10}$$

Divide -2+2\sqrt{31} by 10.

$$x=\frac{\sqrt{31}-1}{5}$$

Now solve the equation x=\frac{-2±2\sqrt{31}}{10} when ± is minus. Subtract 2\sqrt{31} from -2.

$$x=\frac{-2\sqrt{31}-2}{10}$$

Divide -2-2\sqrt{31} by 10.

$$x=\frac{-\sqrt{31}-1}{5}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-1+\sqrt{31}}{5} for x_{1} and \frac{-1-\sqrt{31}}{5} for x_{2}.

$$5x^{2}+2x-6=5\left(x-\frac{\sqrt{31}-1}{5}\right)\left(x-\frac{-\sqrt{31}-1}{5}\right)$$

Evaluate

$$5x^{2}+2x-6$$