Solve 5x^2+4x-6

$$5 { x }^{ 2 } +4x-6$$

Factor

$$5\left(x-\frac{-\sqrt{34}-2}{5}\right)\left(x-\frac{\sqrt{34}-2}{5}\right)$$

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.

$$5x^{2}+4x-6=0$$

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.

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

Square 4.

$$x=\frac{-4±\sqrt{16-4\times 5\left(-6\right)}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-4±\sqrt{16-20\left(-6\right)}}{2\times 5}$$

Multiply -20 times -6.

$$x=\frac{-4±\sqrt{16+120}}{2\times 5}$$

Add 16 to 120.

$$x=\frac{-4±\sqrt{136}}{2\times 5}$$

Take the square root of 136.

$$x=\frac{-4±2\sqrt{34}}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{-4±2\sqrt{34}}{10}$$

Now solve the equation x=\frac{-4±2\sqrt{34}}{10} when ± is plus. Add -4 to 2\sqrt{34}.

$$x=\frac{2\sqrt{34}-4}{10}$$

Divide -4+2\sqrt{34} by 10.

$$x=\frac{\sqrt{34}-2}{5}$$

Now solve the equation x=\frac{-4±2\sqrt{34}}{10} when ± is minus. Subtract 2\sqrt{34} from -4.

$$x=\frac{-2\sqrt{34}-4}{10}$$

Divide -4-2\sqrt{34} by 10.

$$x=\frac{-\sqrt{34}-2}{5}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-2+\sqrt{34}}{5} for x_{1} and \frac{-2-\sqrt{34}}{5} for x_{2}.

$$5x^{2}+4x-6=5\left(x-\frac{\sqrt{34}-2}{5}\right)\left(x-\frac{-\sqrt{34}-2}{5}\right)$$

Evaluate

$$5x^{2}+4x-6$$