What are the solutions to the equation x^2+4x+29=0 ?

The solutions to the equation x^2+4x+29=0 are x=-2+5i,x=-2-5i

Find the zeros of x^2+4x+29=0

The zeros of x^2+4x+29=0 are x=-2+5i,x=-2-5i

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x^{2} + 4 x + 29 = 0

x = - 2 + 5 i, x = - 2 - 5 i

Solution steps

x^{2} + 4 x + 29 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 1 \cdot 29}{2 \cdot 1}

Simplify

4^{2} - 4 \cdot 1 \cdot 29 : 10 i

x_{1, 2} = \frac{- 4 \pm 10 i}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 4 + 10 i}{2 \cdot 1}, x_{2} = \frac{- 4 - 10 i}{2 \cdot 1}

x = \frac{- 4 + 10 i}{2 \cdot 1} : - 2 + 5 i

x = \frac{- 4 - 10 i}{2 \cdot 1} : - 2 - 5 i

The solutions to the quadratic equation are:

x = - 2 + 5 i, x = - 2 - 5 i

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What are the solutions to the equation x^2+4x+29=0 ?
The solutions to the equation x^2+4x+29=0 are x=-2+5i,x=-2-5i
Find the zeros of x^2+4x+29=0
The zeros of x^2+4x+29=0 are x=-2+5i,x=-2-5i