Calculate the value of the discriminant of x^2+x+3=0?

ax² + bx + c = 0

The discriminant (Δ) is b²-4ac

Here:
a = 1
b = 1
c = 3

Δ = 1² - 4×1×3
= 1 - 12
= -11

x^2 + x + 3 = 0
x = [-b ±√(b^2 - 4ac)]/2a

a = 1
b = 1
c = 3

x = [-1 ±√(1 - 12)]/2
x = [-1 ±√-11]/2 (imaginary number)
(no real roots)

x^2+x+3=0

The discriminant is a portion of the quadratic equation, it tells you how many solutions (roots) the equation will have.
The discriminant, D = b^2 - 4ac
where a = coefficient of x^2, b = coefficient of x and c = the constant.
x^2+x+3 ... a = 1, b = 1, c = 3
D = b^2 - 4ac
D= (1)^2 - 4(1)(3) = 1 - 12 = -11

When D > 0, there are two real solutions
When D = 0, there is 1 real root
When D < 0, there are 2 conjugate (imaginary) solutions

Quadratic equation
- b +/- √b² - 4ac / 2a a = 1, b = 1, c = 3
- 1 +/- √(1)² - 4(1)(3) / 2(1) =
- 1 +/- √1 - 12 /2 =
-1 +/- √- 11 / 2 =
-1/2 +√- 11/2, -1/2 - √-11/2

The square root of a negative number is a "conjugate" number, it is part real part imaginary. "i" = √-1, i² = -1

√- 11 = √(-1)(11) =
√(-1) = i √(11) = √11
so √- 11 = i√11
The two imaginary roots are: -1/2 + i√11/2
-1/2 - i√11/2

The discriminant is easy to calculate and is very useful.

I hope all of this made sense. For more help on imaginary (or complex) numbers, check out:
http://www.purplemath.com/modules/complex.htm

b=1
a=1
c=3

b*2-4ac = (1-12) = -11
equation has imaginary roots.

The value of the discriminant is B² - 4AC

1² - 4*1*3 =

1 - 12 =

-11

Here a = 1, b = 1, c = 3

D = b^2 – 4ac = 1 - 12 = -11

Discriminant D = b^2 - 4ac

D = (1^2)- 4(1)(3)
D = -11 <----------- the equation has 2 imaginary roots.

the discriminant is the value of the "mini-function" under the square root of the quadratic equation which is b^2 - 4ac which in this function is 1^2 - 4(1)(3) = 1 - 12 = -11