What are the solutions to the equation x^2+4x-41=0 ?

The solutions to the equation x^2+4x-41=0 are x=-2+3sqrt(5),x=-2-3sqrt(5)

Find the zeros of x^2+4x-41=0

The zeros of x^2+4x-41=0 are x=-2+3sqrt(5),x=-2-3sqrt(5)

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x^{2} + 4 x - 41 = 0

x = - 2 + 35, x = - 2 - 35

Decimal Notation

x = 4.70820 \dots, x = - 8.70820 \dots

Solution steps

x^{2} + 4 x - 41 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 1 \cdot ( - 41 )}{2 \cdot 1}

4^{2} - 4 \cdot 1 \cdot (- 41) = 65

x_{1, 2} = \frac{- 4 \pm 6 5}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 4 + 6 5}{2 \cdot 1}, x_{2} = \frac{- 4 - 6 5}{2 \cdot 1}

x = \frac{- 4 + 6 5}{2 \cdot 1} : - 2 + 35

x = \frac{- 4 - 6 5}{2 \cdot 1} : - 2 - 35

The solutions to the quadratic equation are:

x = - 2 + 35, x = - 2 - 35

\frac{36 - x ^{2}}{x + 6} \geq 0

3 (6 x + 8) = 0

(1 + 2 i)^{3}

(\frac{2}{3} x^{- \frac{1}{3}})^{2}

7 0^{2}

What are the solutions to the equation x^2+4x-41=0 ?
The solutions to the equation x^2+4x-41=0 are x=-2+3sqrt(5),x=-2-3sqrt(5)
Find the zeros of x^2+4x-41=0
The zeros of x^2+4x-41=0 are x=-2+3sqrt(5),x=-2-3sqrt(5)