What are the solutions to the equation x^2+3x+c=0 ?

The solutions to the equation x^2+3x+c=0 are x=(-3+sqrt(9-4c))/2 ,x=(-3-sqrt(9-4c))/2

Find the zeros of x^2+3x+c=0

The zeros of x^2+3x+c=0 are x=(-3+sqrt(9-4c))/2 ,x=(-3-sqrt(9-4c))/2

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x^{2} + 3 x + c = 0

x = \frac{- 3 + 9 - 4 c}{2}, x = \frac{- 3 - 9 - 4 c}{2}

Solution steps

x^{2} + 3 x + c = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 1 \cdot c}{2 \cdot 1}

Simplify

3^{2} - 4 \cdot 1 \cdot c : 9 - 4 c

x_{1, 2} = \frac{- 3 \pm 9 - 4 c}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 3 + 9 - 4 c}{2 \cdot 1}, x_{2} = \frac{- 3 - 9 - 4 c}{2 \cdot 1}

x = \frac{- 3 + 9 - 4 c}{2 \cdot 1} : \frac{- 3 + 9 - 4 c}{2}

x = \frac{- 3 - 9 - 4 c}{2 \cdot 1} : \frac{- 3 - 9 - 4 c}{2}

The solutions to the quadratic equation are:

x = \frac{- 3 + 9 - 4 c}{2}, x = \frac{- 3 - 9 - 4 c}{2}

6 x^{2} + 3 x = 0

2 x^{2} - 9 x - 65 = x^{2} - x

2 x^{2} - 17 x = 9

s^{2} + 3 s + 4 = 0

x, 3 x^{2} + 4 m x + 3 = 0

What are the solutions to the equation x^2+3x+c=0 ?
The solutions to the equation x^2+3x+c=0 are x=(-3+sqrt(9-4c))/2 ,x=(-3-sqrt(9-4c))/2
Find the zeros of x^2+3x+c=0
The zeros of x^2+3x+c=0 are x=(-3+sqrt(9-4c))/2 ,x=(-3-sqrt(9-4c))/2