Question

How do you find the discriminant and how many and what type of solutions does `x^2+3x+7=0` have?

Answer 1

Let's get some theoretical background first.
Assume you have to solve a general quadratic equation
`ax^2+bx+c=0`
You have no recollection of the formula for its solution (like most people). How should you solve it?
Let's apply the logic.

If your equation looks like this
`alpha (x+ beta)^2 + gamma = 0`,
the solution is easy:

1. You subtract `gamma` from both sides, getting
   `alpha(x+beta)^2=-gamma`
2. Divide by `alpha` both sides, getting
   #(x+beta)^2=-gamma/alpha
3. Extract a square root from both side using positive and negative values of a square root, getting
   `x+beta=sqrt(-gamma/alpha)`
4. Subtract `beta` from both sides, getting
   `x_(1,2)=+-sqrt(-gamma/alpha)-beta`

Our general equation is not of the form we can easily solve, as above, but can be transformed into this form.
Let's equate two forms and find the necessary `alpha, beta, gamma` coefficients:
`ax^2+bx+c=alpha(x+beta)^2+gamma`

Open the parenthesis on the right:
`ax^2+bx+c=alpha x^2 + 2 alpha beta x + alpha beta^2 + gamma`

Now you can determine `alpha, beta, gamma` coefficients by comparing the coefficient at `x` in different degrees:
`a = alpha` (coefficient at `x^2`)
`b = 2 alpha beta` (coefficient at `x`)
`c = alpha beta ^2 + gamma` (free coefficient)
from which follows:
`alpha = a`
`beta = b/(2 alpha) = b/(2a)`
`gamma = c - alpha beta ^2 = c - a b^2/(4a^2) = c - b^2/(4a) = (4ac-b^2)/(4a)`

Now we use the formula that expressed the solution in terms of `alpha, beta, gamma`, substituting these coefficients with their representation in terms of `a, b, c`.

`x_(1,2)=+-sqrt(-gamma/alpha)-beta = +=sqrt((-4ac+b^2)/(4a))-b/(2a) = [-b+-sqrt(b^2-4ac)]/(2a)`

The most involved part of this expression is under the square root, it's called discriminant of this quadratic equation:
`D = b^2-4ac`

If `D=b^2-4ac>0`, there are two distinct real solution to this quadratic equation because the square root of the positive number has two distinct real values.

If `D=b^2-4ac=0`, there is only one real solution since the square root equals to 0 and our `+-` operation renders always `0`.

Finally, if `D=b^2-4ac < 0`, there are no real solution because there are no real values for a square root of a negative number.

In a particular case of this problem with an equation
`x^2+3x+7=0`
the coefficients are:
`a=1, b=3, c=7`
The discriminant `D=b^2-4ac=3^2-4*1*7=-19`
It's negative. Therefore, there are no real solutions.

A more detailed explanation of topics touched in this answer can be found on Unizor by following the menu items Algebra - Quadratic Equations.