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

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

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

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

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

c = - 3, c = - 8

Solution steps

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

Solve with the quadratic formula

c_{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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

c = - 3, c = - 8

8 x - 4 = \frac{8 x + 5}{2}

\frac{x ^{\frac{1}{2}}}{x ^{\frac{5}{8}}}

4 m + 3 - m = 3 (m + 1)

54 x^{2} y - 6 y

∣ x - 7 ∣ < - 5

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