# How do you find the roots, real and imaginary, of y= 3x^2-5x- 1  using the quadratic formula?

Jan 29, 2016

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

#### Explanation:

$y = 3 {x}^{2} - 5 x - 1$ is a quadratic equation in standard form, $y = a x + b x + c$, where $a = 3 , b = - 5 , c = - 1$, and $y = 0$.

The roots are the solutions for $x$.

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

Substitute the known values into the equation.

$x = \frac{- \left(- 5\right) \pm \sqrt{{\left(- 5\right)}^{2} - \left(4 \cdot 3 \cdot - 1\right)}}{2 \cdot 3}$

Simplify.

$x = \frac{5 \pm \sqrt{25 + 12}}{6}$

Add $25$ and $12$.

$x = \frac{5 \pm \sqrt{27}}{6}$

Simplify $\sqrt{27}$.

$\sqrt{27} = \sqrt{3 \cdot 3 \cdot 3} = \sqrt{{3}^{2} \cdot 3} = 3 \sqrt{3}$

$x = \frac{5 \pm 3 \sqrt{3}}{6}$

Write the value for $x$ as two fractions.

$x = \frac{5}{6} \pm \frac{3 \sqrt{3}}{6}$

Simplify $\frac{3}{6}$ to $\frac{1}{2}$.

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