# How do you find the discriminant and how many and what type of solutions does 6x^2=13x have?

May 10, 2015

Ripan's solutions are correct but don't answer the more general question.

Solutions to a quadratic of the form $a {x}^{2} + b x + c = 0$ are given by
the quadratic formula $\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

The sub-expression within the square root determines the number (and type) of solutions; this sub-expression is called the "discriminant" and is typically expressed as:
$\Delta = {b}^{2} - 4 a c$
with the conditions
$\Delta \left\{\begin{matrix}< 0 \text{ there are no Real solutions" \\ =0" there is 1 Real solution" \\ >0" there are 2 Real solutions}\end{matrix}\right.$

Given $6 {x}^{2} = 13 x$
we can re-arrange this into the general form
$6 {x}^{2} - 13 x + 0 = 0$

and
$\Delta = {\left(13\right)}^{2} - 4 \left(6\right) \left(0\right) = 169 > 0$
so $6 {x}^{2} = 13 x$ has 2 Real solutions