How to know how much this cubic equation have real answers?

Question

I have been hours trying to solve this cubic equation any help? $2x^3-7x^2+1.6x+7=0$

Answer 1

If the qubic equation $$ax^3+bx^2+cx+d=0\tag{1}$$ is given, then one of easiest ways to know the number of its real roots (since the number of complex roots is $3$ anyway) is to calculate special value called Discriminant (Discriminant (Wiki)).

It is some sort of "indicators" of the number of real roots for polynomials of small degree.

For quadratic equations $ax^2+bx+c=0$ there exists simple formula: $D=b^2-4ac$.

For cubic equation of the form $(1)$ we have a bit longer formula:

$$ D = b^2c^2 - 4ac^3 - 4b^3d - 27a^2d^2+18abcd. \tag{2} $$

If $D=0$, then at least two solutions are equal.

If $D<0$, then there is $1$ real solution of eq. ($1$) (and two complex conjugate ones).

If $D>0$, then there are three distinct real solutions.

When evaluate expression $(2)$, one will obtain positive value, so?

Answer 2

To solve this question, you can use the cubic formula, it is a very long and complicated formula, I hope this helps!