# How do you find the real or imaginary solutions of the equation 6x^2+13x-5=0?

Nov 28, 2016

The solutions are $S = \left\{\frac{1}{3} , - \frac{5}{2}\right\}$

#### Explanation:

The simultaneous equations $a {x}^{2} + b x + c = 0$

Our equation is $6 {x}^{2} + 13 x - 5 = 0$

We calculate the determinant,

$\Delta = {b}^{2} - 4 a c$

$\Delta = {13}^{2} - 4 \cdot 6 \cdot - 5 = 289$

$\Delta > 0$, so we have 2 real roots

So,

$x = \frac{- b \pm \sqrt{\Delta}}{2 a}$

$= \frac{- 13 \pm \sqrt{289}}{12}$

$= \frac{- 13 \pm 17}{12}$

So, ${x}_{1} = - \frac{30}{12} = - \frac{5}{2}$

and ${x}_{2} = \frac{4}{12} = \frac{1}{3}$