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

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

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

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

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

r = - 3, r = - 8

Solution steps

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

Solve with the quadratic formula

r_{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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

r = - 3, r = - 8

2 x^{2} = - 30 x - 112

(m + 3) (m + 2) = 0

(2 x + 3)^{2} - 4 x = 13 - 7 x

- 4 x^{2} + 32 x + 80 = 0

x^{2} + 8 x = - 12 + 0

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