What are the solutions to the equation x^2+4x+52=0 ?

The solutions to the equation x^2+4x+52=0 are x=-2+4sqrt(3)i,x=-2-4sqrt(3)i

Find the zeros of x^2+4x+52=0

The zeros of x^2+4x+52=0 are x=-2+4sqrt(3)i,x=-2-4sqrt(3)i

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x^{2} + 4 x + 52 = 0

x = - 2 + 43 i, x = - 2 - 43 i

Solution steps

x^{2} + 4 x + 52 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 1 \cdot 52}{2 \cdot 1}

Simplify

4^{2} - 4 \cdot 1 \cdot 52 : 83 i

x_{1, 2} = \frac{- 4 \pm 8 3 i}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 4 + 8 3 i}{2 \cdot 1}, x_{2} = \frac{- 4 - 8 3 i}{2 \cdot 1}

x = \frac{- 4 + 8 3 i}{2 \cdot 1} : - 2 + 43 i

x = \frac{- 4 - 8 3 i}{2 \cdot 1} : - 2 - 43 i

The solutions to the quadratic equation are:

x = - 2 + 43 i, x = - 2 - 43 i

x, x^{2} + y^{2} = 2 a x

0 = x^{2} - 5 x - 36

14 (x + 1) (x + 2) - 18 (x - 2) (x + 2) = 22 (x - 2)

x^{2} + 1.8 x - 0.225 = 0

x = (2 - x)^{2}

What are the solutions to the equation x^2+4x+52=0 ?
The solutions to the equation x^2+4x+52=0 are x=-2+4sqrt(3)i,x=-2-4sqrt(3)i
Find the zeros of x^2+4x+52=0
The zeros of x^2+4x+52=0 are x=-2+4sqrt(3)i,x=-2-4sqrt(3)i