What are the solutions to the equation x^2+12x-3=0 ?

The solutions to the equation x^2+12x-3=0 are x=-6+sqrt(39),x=-6-sqrt(39)

Find the zeros of x^2+12x-3=0

The zeros of x^2+12x-3=0 are x=-6+sqrt(39),x=-6-sqrt(39)

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x^{2} + 12 x - 3 = 0

x = - 6 + 39, x = - 6 - 39

Solution steps

x^{2} + 12 x - 3 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 1 \cdot ( - 3 )}{2 \cdot 1}

1 2^{2} - 4 \cdot 1 \cdot (- 3) = 239

x_{1, 2} = \frac{- 12 \pm 2 39}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 12 + 2 39}{2 \cdot 1}, x_{2} = \frac{- 12 - 2 39}{2 \cdot 1}

x = \frac{- 12 + 2 39}{2 \cdot 1} : - 6 + 39

x = \frac{- 12 - 2 39}{2 \cdot 1} : - 6 - 39

The solutions to the quadratic equation are:

x = - 6 + 39, x = - 6 - 39

12 3 - 7108

\frac{3 x}{5} - \frac{x - 3}{2} = \frac{x + 2}{3}

\frac{\frac{7}{6}}{8}

6 x^{2} - 18 x - 60

3 x - 5 \geq 13

What are the solutions to the equation x^2+12x-3=0 ?
The solutions to the equation x^2+12x-3=0 are x=-6+sqrt(39),x=-6-sqrt(39)
Find the zeros of x^2+12x-3=0
The zeros of x^2+12x-3=0 are x=-6+sqrt(39),x=-6-sqrt(39)