What are the solutions to the equation a^2+11a+24=0 ?

The solutions to the equation a^2+11a+24=0 are a=-3,a=-8

Find the zeros of a^2+11a+24=0

The zeros of a^2+11a+24=0 are a=-3,a=-8

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a^{2} + 11 a + 24 = 0

a = - 3, a = - 8

Solution steps

a^{2} + 11 a + 24 = 0

Solve with the quadratic formula

a_{1, 2} = \frac{- 11 \pm 1 1 ^{2} - 4 \cdot 1 \cdot 24}{2 \cdot 1}

1 1^{2} - 4 \cdot 1 \cdot 24 = 5

a_{1, 2} = \frac{- 11 \pm 5}{2 \cdot 1}

Separate the solutions

a_{1} = \frac{- 11 + 5}{2 \cdot 1}, a_{2} = \frac{- 11 - 5}{2 \cdot 1}

a = \frac{- 11 + 5}{2 \cdot 1} : - 3

a = \frac{- 11 - 5}{2 \cdot 1} : - 8

The solutions to the quadratic equation are:

a = - 3, a = - 8

7 x = 4 x^{2} - 1

(u - 2)^{2} - 75 = 0

9 x^{2} + 6 x + 1 = 0

(m^{2} + 1) = 0

2 - 3 x^{2} = 0

What are the solutions to the equation a^2+11a+24=0 ?
The solutions to the equation a^2+11a+24=0 are a=-3,a=-8
Find the zeros of a^2+11a+24=0
The zeros of a^2+11a+24=0 are a=-3,a=-8