Solve 3x^2-3x-5

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

Factor

$$3\left(x-\left(-\frac{\sqrt{69}}{6}+\frac{1}{2}\right)\right)\left(x-\left(\frac{\sqrt{69}}{6}+\frac{1}{2}\right)\right)$$

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.

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

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.

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

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\times 3\left(-5\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-3\right)±\sqrt{9-12\left(-5\right)}}{2\times 3}$$

Multiply -12 times -5.

$$x=\frac{-\left(-3\right)±\sqrt{9+60}}{2\times 3}$$

Add 9 to 60.

$$x=\frac{-\left(-3\right)±\sqrt{69}}{2\times 3}$$

The opposite of -3 is 3.

$$x=\frac{3±\sqrt{69}}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{3±\sqrt{69}}{6}$$

Now solve the equation x=\frac{3±\sqrt{69}}{6} when ± is plus. Add 3 to \sqrt{69}.

$$x=\frac{\sqrt{69}+3}{6}$$

Divide 3+\sqrt{69} by 6.

$$x=\frac{\sqrt{69}}{6}+\frac{1}{2}$$

Now solve the equation x=\frac{3±\sqrt{69}}{6} when ± is minus. Subtract \sqrt{69} from 3.

$$x=\frac{3-\sqrt{69}}{6}$$

Divide 3-\sqrt{69} by 6.

$$x=-\frac{\sqrt{69}}{6}+\frac{1}{2}$$

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{69}}{6} for x_{1} and \frac{1}{2}-\frac{\sqrt{69}}{6} for x_{2}.

$$3x^{2}-3x-5=3\left(x-\left(\frac{\sqrt{69}}{6}+\frac{1}{2}\right)\right)\left(x-\left(-\frac{\sqrt{69}}{6}+\frac{1}{2}\right)\right)$$

Evaluate

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