# How to use the discriminant to find out what type of solutions the equation has for 6-5x=6x^2?

May 23, 2015

First add $5 x - 6$ to both sides to get:

$6 {x}^{2} + 5 x - 6 = 0$

This is of the form $a {x}^{2} + b x + c = 0$, with $a = 6$, $b = 5$ and $c = - 6$ so the discriminant is given by the formula:

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

$= {5}^{2} - \left(4 \times 6 \times - 6\right) = 25 + 144 = 169 = {13}^{2}$

That $\Delta$ is positive means that the equation has two distinct real solutions. That $\Delta$ is a perfect square means that the solutions are rational.

The other possible cases are:

$\Delta = 0$ The equation has one repeated, real, rational root.

$\Delta < 0$ The equation has no real roots. It has two distinct complex roots (which are complex conjugates).

In our case, the solutions are given by the formula:

$x = \frac{- b \pm \sqrt{\Delta}}{2 a} = \frac{- 5 \pm 13}{12}$

That is $x = - \frac{18}{12} = - \frac{3}{2}$ or $x = \frac{8}{12} = \frac{2}{3}$

This also means that $6 {x}^{2} + 5 x - 6$

can be factored as $\left(2 x + 3\right) \left(3 x - 2\right)$