Solve 5x^2+6x-3

$$5 { x }^{ 2 } +6x-3$$

Factor

$$5\left(x-\frac{-2\sqrt{6}-3}{5}\right)\left(x-\frac{2\sqrt{6}-3}{5}\right)$$

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.

$$5x^{2}+6x-3=0$$

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.

$$x=\frac{-6±\sqrt{6^{2}-4\times 5\left(-3\right)}}{2\times 5}$$

Square 6.

$$x=\frac{-6±\sqrt{36-4\times 5\left(-3\right)}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-6±\sqrt{36-20\left(-3\right)}}{2\times 5}$$

Multiply -20 times -3.

$$x=\frac{-6±\sqrt{36+60}}{2\times 5}$$

Add 36 to 60.

$$x=\frac{-6±\sqrt{96}}{2\times 5}$$

Take the square root of 96.

$$x=\frac{-6±4\sqrt{6}}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{-6±4\sqrt{6}}{10}$$

Now solve the equation x=\frac{-6±4\sqrt{6}}{10} when ± is plus. Add -6 to 4\sqrt{6}.

$$x=\frac{4\sqrt{6}-6}{10}$$

Divide -6+4\sqrt{6} by 10.

$$x=\frac{2\sqrt{6}-3}{5}$$

Now solve the equation x=\frac{-6±4\sqrt{6}}{10} when ± is minus. Subtract 4\sqrt{6} from -6.

$$x=\frac{-4\sqrt{6}-6}{10}$$

Divide -6-4\sqrt{6} by 10.

$$x=\frac{-2\sqrt{6}-3}{5}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-3+2\sqrt{6}}{5} for x_{1} and \frac{-3-2\sqrt{6}}{5} for x_{2}.

$$5x^{2}+6x-3=5\left(x-\frac{2\sqrt{6}-3}{5}\right)\left(x-\frac{-2\sqrt{6}-3}{5}\right)$$

Evaluate

$$5x^{2}+6x-3$$