What are the solutions to the equation roots x^2+x+c ?

The solutions to the equation roots x^2+x+c are x=(-1+sqrt(1-4c))/2 ,x=(-1-sqrt(1-4c))/2

Find the zeros of roots x^2+x+c

The zeros of roots x^2+x+c are x=(-1+sqrt(1-4c))/2 ,x=(-1-sqrt(1-4c))/2

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roots

x^{2} + x + c

x = \frac{- 1 + 1 - 4 c}{2}, x = \frac{- 1 - 1 - 4 c}{2}

Solution steps

x^{2} + x + c

The roots are the intercepts with the x-axis

(y = 0)

x^{2} + x + c = 0

Solve with the quadratic formula

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

Simplify

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = \frac{- 1 + 1 - 4 c}{2}, x = \frac{- 1 - 1 - 4 c}{2}

6^{2} = x^{2} + x^{2}

x, x^{2} + y^{2} + z^{2} = z

\frac{x ^{2}}{25} = 0.5

x, 0 = 12 x^{2} - 96 x + 135

z^{2} = - 12

What are the solutions to the equation roots x^2+x+c ?
The solutions to the equation roots x^2+x+c are x=(-1+sqrt(1-4c))/2 ,x=(-1-sqrt(1-4c))/2
Find the zeros of roots x^2+x+c
The zeros of roots x^2+x+c are x=(-1+sqrt(1-4c))/2 ,x=(-1-sqrt(1-4c))/2