Question

How many real number solutions? `-7x^2+6x+3=0`

Answer 1

discriminant = 120

two real irrational solutions.

Explanation:

`-7x^2+6x+3`

To solve this use the discriminant:

`ax^2+bx+c`

discriminant `=b^2-4ac`

`=6^2-4*-7*3`

`=36-(-84)`

`=120`

discriminant > 0: two real solutions.

discriminant = 0: one real solutions, bounce or double solution.

discriminant < 0: two imaginary solutions.

discriminant = perfect square: solution is rational

Answer 2

`d = 120`, therefore, there are two distinct real roots.

Explanation:

One can find the number of real solutions of a quadratic of the form, `y = ax^2+bx+c`, by computing the determinant, `d = b^2-4ac`

If `d < 0`, then there are no real roots.
If `d=0`, then there is one real root (called a repeated root).
If `d >0`, then there are two distinct real roots:

Given: `-7x^2+6x+3=0`

`d = 6^2-4(-7)(3)`

`d = 120`, therefore, there are two distinct real roots.