What are the solutions to the equation c^2+5c+4=0 ?

The solutions to the equation c^2+5c+4=0 are c=-1,c=-4

Find the zeros of c^2+5c+4=0

The zeros of c^2+5c+4=0 are c=-1,c=-4

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c^{2} + 5 c + 4 = 0

c = - 1, c = - 4

Solution steps

c^{2} + 5 c + 4 = 0

Solve with the quadratic formula

c_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 1 \cdot 4}{2 \cdot 1}

5^{2} - 4 \cdot 1 \cdot 4 = 3

c_{1, 2} = \frac{- 5 \pm 3}{2 \cdot 1}

Separate the solutions

c_{1} = \frac{- 5 + 3}{2 \cdot 1}, c_{2} = \frac{- 5 - 3}{2 \cdot 1}

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

c = \frac{- 5 - 3}{2 \cdot 1} : - 4

The solutions to the quadratic equation are:

c = - 1, c = - 4

k^{2} - 25 = 0

7 x^{2} - 47 x - 14 = 0

\frac{- n ^{2} - 4}{2} = 16

0.036 = (1 - V^{2})

2 y^{2} - 3 = 5 y

What are the solutions to the equation c^2+5c+4=0 ?
The solutions to the equation c^2+5c+4=0 are c=-1,c=-4
Find the zeros of c^2+5c+4=0
The zeros of c^2+5c+4=0 are c=-1,c=-4