How do you solve 3x^2-4x-5=0 using the quadratic formula?

Apr 5, 2018

$x = \frac{2 + \sqrt{19}}{3} ,$ $\frac{2 - \sqrt{19}}{3}$

Explanation:

Solve:

$3 {x}^{2} - 4 x - 5 = 0$ is a quadratic equation in standard form:

$a {x}^{2} + b x + c$,

where:

$a = 3$, $b = - 4$, $c = - 5$

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

Plug in the known values.

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

Simplify.

$x = \frac{4 \pm \sqrt{76}}{6}$

Prime factorize $76$.

$x = \frac{4 \pm \sqrt{{2}^{2} \times 19}}{6}$

Apply rule: $\sqrt{{a}^{2}} = a$

$x = \frac{4 \pm 2 \sqrt{19}}{6}$

Simplify.

$x = \frac{2 \pm \sqrt{19}}{3}$

Solutions for $x$.

$x = \frac{2 + \sqrt{19}}{3} ,$ $\frac{2 - \sqrt{19}}{3}$