# How do you solve  3x^2 - 14x - 5 = 0?

Aug 3, 2015

${x}_{1 , 2} = \frac{14 \pm 16}{6}$

#### Explanation:

For a quadratic equation that takes the general form

$\textcolor{b l u e}{a {x}^{2} + b x + c = 0}$

the two solutions can be determined by using the quadratic formula

color(blue)(x_(1,2) = (-b +- sqrt(b^2 - 4ac))/(2a)

In your case, $a = 3$, $b = - 14$, and $c = - 5$. This means that the two solutions will be

${x}_{1 , 2} = \frac{- \left(- 14\right) \pm \sqrt{{\left(- 14\right)}^{2} - 4 \cdot 3 \cdot \left(- 5\right)}}{2 \cdot 3}$

${x}_{1 , 2} = \frac{14 \pm \sqrt{256}}{6}$

${x}_{1 , 2} = \frac{14 \pm 16}{6} = \left\{\begin{matrix}{x}_{1} = \frac{14 + 16}{6} = \textcolor{g r e e n}{5} \\ {x}_{2} = \frac{14 - 16}{6} = \textcolor{g r e e n}{- \frac{1}{3}}\end{matrix}\right.$

Aug 4, 2015

Solve y = 3x^2 - 14x - 5 = 0 (1)

Ans: (- 1/3) and (5).

#### Explanation:

I use the new Transforming Method.
Transformed y' = x^2 - 14x - 15 = 0. (2)
Since (a -b + c = 0) --> 2 real roots of (2) are (-1) and (-c/a = 15).
Back to equation (1), the 2 real roots are: $\left(- \frac{1}{3}\right)$ and $\left(\frac{15}{3} = 5\right)$