Question

How do you find the discriminant and how many solutions does `x^2+4x+3=4` have?

Answer 1

We get two real solutions

Explanation:

Writing your equation in the form
`x^2+4x-1=0`
the4n we get by the quadratic formula:

`x_{1,2}=2pm\sqrt(4+1)`
so we get
`x_1=2+sqrt(5)`

`x_2=2-sqrt(5)`

Answer 2

`x = -2(1 +- sqrt5)`

Explanation:

`x^2 + 4x + 3 = 4`
`f(x) = x^2 + 4x - 1 = 0`
Discriminant D -->
`D = d^2 = b^2 - 4ac = 16 + 4 = 20` --> `d = +- 4sqrt5`
There are 2 real roots:
`x = -b/(2a) +- d/(2a) = -4/2 +- (4sqrt5)/2 = - 2 +- 2sqrt5`