x^2+11x+30>= 0 — step-by-step solution

Let's solve in

ℝ

x^{2} + 11 x + 30 \geq 0

Given the quadratic form

a x^{2} + b x + c

, you can calculate the discriminant:

Δ = b^{2} - 4 a c

. Here

a = 1

b = 11

and

c = 30

Δ = 11^{2} - 4 \cdot 1 \cdot 30

ⓘ Powers are computed before multiplications, additions and subtractions. An integer power is a repeated product:

11^{2}

11 \cdot 11

121

Δ = 121 - 4 \cdot 30

ⓘ Multiplications are done before additions and subtractions:

4 \cdot 30

120

Δ = 121 - 120

ⓘ Start from

121

and move back

120

on the number line:

121 - 120

1

Δ = 1

Since

Δ > 0

, the equation has two roots:

The quadratic formula gives

x_{1}

\frac{- b - \sqrt{Δ}}{2 a}

\frac{- 11 - \sqrt{1}}{2 \cdot 1}

x_{1} = \frac{- 11 - 1}{2}

ⓘ The numerator and the denominator are computed before the division:

- 11 - 1

-12

x_{1} = - \frac{12}{2}

ⓘ We simplify the fraction by dividing the numerator and the denominator by

2

, their greatest common divisor:

- \frac{12}{2}

-6

x_{1} = -6

The quadratic formula gives

x_{2}

\frac{- b + \sqrt{Δ}}{2 a}

\frac{- 11 + \sqrt{1}}{2 \cdot 1}

x_{2} = \frac{- 11 + 1}{2}

ⓘ The numerator and the denominator are computed before the division:

- 11 + 1

-10

x_{2} = - \frac{10}{2}

ⓘ We simplify the fraction by dividing the numerator and the denominator by

2

, their greatest common divisor:

- \frac{10}{2}

-5

x_{2} = -5

The quadratic has the sign of a =

1

outside its roots, so it is indeed positive outside its roots; the roots, where it is zero, are included.

▶Solution obtained:

x \in (-∞, -6] \cup [-5, +∞)