Question

What is the solution of 5=17x^2-200x?

(A) 1 real number (B) 2 complex numbers that are conjugates (C) 2 real unequal numbers (D) ∅ (E) none of these

Answer 1

lazy lazy lazy http://openstudy.com/study#/updates/550b0d3ee4b0f4dae08515e9

Answer 2

$$\Large {17x^2-200x=5\\
\\ \iff \\

17x^2 - 200x - 5 = 0

}
$$

Answer 3

but you can look at the discriminant to find how many roots, what type of roots

Answer 4

If you've already asked this question, and you received help, please don't ask the question again in a new tab. We're not going to do all the work for you.

Answer 5

$$\Large {
\text {Given: }~ax^2+bx+c = 0\\
\\ \text{There are three cases}\\
\begin{cases}
~b^2-4ac >0 & then~ 2~real ~unequal ~roots \\
~b^2-4ac =0 & then~ 1~real ~root~ repeated \\
b^2-4ac <0 & then~ 2~imaginary~ roots
\end{cases}

}
$$