Solve 5x^2+5x-6

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

Factor

$$5\left(x-\left(-\frac{\sqrt{145}}{10}-\frac{1}{2}\right)\right)\left(x-\left(\frac{\sqrt{145}}{10}-\frac{1}{2}\right)\right)$$

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.

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

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.

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

Square 5.

$$x=\frac{-5±\sqrt{25-4\times 5\left(-6\right)}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-5±\sqrt{25-20\left(-6\right)}}{2\times 5}$$

Multiply -20 times -6.

$$x=\frac{-5±\sqrt{25+120}}{2\times 5}$$

Add 25 to 120.

$$x=\frac{-5±\sqrt{145}}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{-5±\sqrt{145}}{10}$$

Now solve the equation x=\frac{-5±\sqrt{145}}{10} when ± is plus. Add -5 to \sqrt{145}.

$$x=\frac{\sqrt{145}-5}{10}$$

Divide -5+\sqrt{145} by 10.

$$x=\frac{\sqrt{145}}{10}-\frac{1}{2}$$

Now solve the equation x=\frac{-5±\sqrt{145}}{10} when ± is minus. Subtract \sqrt{145} from -5.

$$x=\frac{-\sqrt{145}-5}{10}$$

Divide -5-\sqrt{145} by 10.

$$x=-\frac{\sqrt{145}}{10}-\frac{1}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{1}{2}+\frac{\sqrt{145}}{10} for x_{1} and -\frac{1}{2}-\frac{\sqrt{145}}{10} for x_{2}.

$$5x^{2}+5x-6=5\left(x-\left(\frac{\sqrt{145}}{10}-\frac{1}{2}\right)\right)\left(x-\left(-\frac{\sqrt{145}}{10}-\frac{1}{2}\right)\right)$$

Evaluate

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