What are the solutions to the equation roots x^2+12x+c ?

The solutions to the equation roots x^2+12x+c are x=-6+sqrt(-c+36),x=-sqrt(-c+36)-6

Find the zeros of roots x^2+12x+c

The zeros of roots x^2+12x+c are x=-6+sqrt(-c+36),x=-sqrt(-c+36)-6

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roots

x^{2} + 12 x + c

x = - 6 + - c + 36, x = - - c + 36 - 6

Solution steps

x^{2} + 12 x + c

The roots are the intercepts with the x-axis

(y = 0)

x^{2} + 12 x + c = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 1 \cdot c}{2 \cdot 1}

Simplify

1 2^{2} - 4 \cdot 1 \cdot c : 2 36 - c

x_{1, 2} = \frac{- 12 \pm 2 36 - c}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 12 + 2 36 - c}{2 \cdot 1}, x_{2} = \frac{- 12 - 2 36 - c}{2 \cdot 1}

x = \frac{- 12 + 2 36 - c}{2 \cdot 1} : - 6 + - c + 36

x = \frac{- 12 - 2 36 - c}{2 \cdot 1} : - - c + 36 - 6

The solutions to the quadratic equation are:

x = - 6 + - c + 36, x = - - c + 36 - 6

- x (- 6 (x) + 1) = 1

2 a^{2} + 4 a - 6 = - 3 a

y, 2 x^{2} - x y + y^{2} = 16

y, x^{2} = \frac{1}{y} - 1

x^{2} + \frac{2}{3} x - \frac{1}{9} = 0

What are the solutions to the equation roots x^2+12x+c ?
The solutions to the equation roots x^2+12x+c are x=-6+sqrt(-c+36),x=-sqrt(-c+36)-6
Find the zeros of roots x^2+12x+c
The zeros of roots x^2+12x+c are x=-6+sqrt(-c+36),x=-sqrt(-c+36)-6