What are the solutions to the equation x^2+4x+36=0 ?

The solutions to the equation x^2+4x+36=0 are x=-2+4sqrt(2)i,x=-2-4sqrt(2)i

Find the zeros of x^2+4x+36=0

The zeros of x^2+4x+36=0 are x=-2+4sqrt(2)i,x=-2-4sqrt(2)i

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x^{2} + 4 x + 36 = 0

x = - 2 + 42 i, x = - 2 - 42 i

Solution steps

x^{2} + 4 x + 36 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 1 \cdot 36}{2 \cdot 1}

Simplify

4^{2} - 4 \cdot 1 \cdot 36 : 82 i

x_{1, 2} = \frac{- 4 \pm 8 2 i}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 4 + 8 2 i}{2 \cdot 1}, x_{2} = \frac{- 4 - 8 2 i}{2 \cdot 1}

x = \frac{- 4 + 8 2 i}{2 \cdot 1} : - 2 + 42 i

x = \frac{- 4 - 8 2 i}{2 \cdot 1} : - 2 - 42 i

The solutions to the quadratic equation are:

x = - 2 + 42 i, x = - 2 - 42 i

- (n - 8) = - 2

2601

∣ 2 x - 7 ∣ < 1

4 x^{2} - 36 x + 80 = 0

172

What are the solutions to the equation x^2+4x+36=0 ?
The solutions to the equation x^2+4x+36=0 are x=-2+4sqrt(2)i,x=-2-4sqrt(2)i
Find the zeros of x^2+4x+36=0
The zeros of x^2+4x+36=0 are x=-2+4sqrt(2)i,x=-2-4sqrt(2)i