Solve (5x^2-4x-3)

$$(5 { x }^{ 2 } -4x-3)$$

Factor

$$5\left(x-\frac{2-\sqrt{19}}{5}\right)\left(x-\frac{\sqrt{19}+2}{5}\right)$$

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.

$$5x^{2}-4x-3=0$$

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.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 5\left(-3\right)}}{2\times 5}$$

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\times 5\left(-3\right)}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-\left(-4\right)±\sqrt{16-20\left(-3\right)}}{2\times 5}$$

Multiply -20 times -3.

$$x=\frac{-\left(-4\right)±\sqrt{16+60}}{2\times 5}$$

Add 16 to 60.

$$x=\frac{-\left(-4\right)±\sqrt{76}}{2\times 5}$$

Take the square root of 76.

$$x=\frac{-\left(-4\right)±2\sqrt{19}}{2\times 5}$$

The opposite of -4 is 4.

$$x=\frac{4±2\sqrt{19}}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{4±2\sqrt{19}}{10}$$

Now solve the equation x=\frac{4±2\sqrt{19}}{10} when ± is plus. Add 4 to 2\sqrt{19}.

$$x=\frac{2\sqrt{19}+4}{10}$$

Divide 4+2\sqrt{19} by 10.

$$x=\frac{\sqrt{19}+2}{5}$$

Now solve the equation x=\frac{4±2\sqrt{19}}{10} when ± is minus. Subtract 2\sqrt{19} from 4.

$$x=\frac{4-2\sqrt{19}}{10}$$

Divide 4-2\sqrt{19} by 10.

$$x=\frac{2-\sqrt{19}}{5}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{2+\sqrt{19}}{5} for x_{1} and \frac{2-\sqrt{19}}{5} for x_{2}.

$$5x^{2}-4x-3=5\left(x-\frac{\sqrt{19}+2}{5}\right)\left(x-\frac{2-\sqrt{19}}{5}\right)$$

Evaluate

$$5x^{2}-4x-3$$