What are the solutions to the equation x^2+12x-5=0 ?

The solutions to the equation x^2+12x-5=0 are x=-6+sqrt(41),x=-6-sqrt(41)

Find the zeros of x^2+12x-5=0

The zeros of x^2+12x-5=0 are x=-6+sqrt(41),x=-6-sqrt(41)

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x^{2} + 12 x - 5 = 0

x = - 6 + 41, x = - 6 - 41

Solution steps

x^{2} + 12 x - 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 1 \cdot ( - 5 )}{2 \cdot 1}

1 2^{2} - 4 \cdot 1 \cdot (- 5) = 241

x_{1, 2} = \frac{- 12 \pm 2 41}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 12 + 2 41}{2 \cdot 1}, x_{2} = \frac{- 12 - 2 41}{2 \cdot 1}

x = \frac{- 12 + 2 41}{2 \cdot 1} : - 6 + 41

x = \frac{- 12 - 2 41}{2 \cdot 1} : - 6 - 41

The solutions to the quadratic equation are:

x = - 6 + 41, x = - 6 - 41

(\frac{4}{7})^{1}

9 x (a - b) - 2 y (a - b)

t^{4} - p^{4}

- 4 ∣ x ∣ + 10 = - 10

\frac{3 ( x + 1 )}{4} = 2 x - 3

What are the solutions to the equation x^2+12x-5=0 ?
The solutions to the equation x^2+12x-5=0 are x=-6+sqrt(41),x=-6-sqrt(41)
Find the zeros of x^2+12x-5=0
The zeros of x^2+12x-5=0 are x=-6+sqrt(41),x=-6-sqrt(41)