## anonymous one year ago Where is the removable discontinuity of f(x)=(x+5)/(2x^2+3x-10) located a.x=-5 b.x = –2 c.x = 2 d.x = 5

1. Michele_Laino

please can you check your formula of f(x)?

2. Michele_Laino

is the denominator, like this: $2{x^2} + 3x - 10$

3. anonymous

yes

4. Michele_Laino

the solution of this equation: $2{x^2} + 3x - 10 = 0$ are: $\begin{gathered} x = \frac{{ - 3 \pm \sqrt {{3^2} - 4 \times 2 \times \left( { - 10} \right)} }}{{2 \times 2}} = \hfill \\ \hfill \\ = \frac{{ - 3 \pm \sqrt {9 + 80} }}{4} = \frac{{ - 3 \pm \sqrt {89} }}{4} \hfill \\ \end{gathered}$ am I right?

5. Michele_Laino

I have applied the standard formula: $x = \frac{{ - b \pm \sqrt {{b^2} - 4 \times a \times c} }}{{2 \times a}}$

6. Michele_Laino

I think that the denominator can be this: ${x^2} + 3x - 10$ is it possible?