What are the solutions to the equation 0=x^2+4x+16 ?

The solutions to the equation 0=x^2+4x+16 are x=-2+2sqrt(3)i,x=-2-2sqrt(3)i

Find the zeros of 0=x^2+4x+16

The zeros of 0=x^2+4x+16 are x=-2+2sqrt(3)i,x=-2-2sqrt(3)i

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0 = x^{2} + 4 x + 16

x = - 2 + 23 i, x = - 2 - 23 i

Solution steps

0 = x^{2} + 4 x + 16

Switch sides

x^{2} + 4 x + 16 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 1 \cdot 16}{2 \cdot 1}

Simplify

4^{2} - 4 \cdot 1 \cdot 16 : 43 i

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

Separate the solutions

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

x = \frac{- 4 + 4 3 i}{2 \cdot 1} : - 2 + 23 i

x = \frac{- 4 - 4 3 i}{2 \cdot 1} : - 2 - 23 i

The solutions to the quadratic equation are:

x = - 2 + 23 i, x = - 2 - 23 i

x^{2} - 4 x = 3 x + 8

(x + 3)^{2} - 3 = (2 x + 2)^{2}

X^{2} - 1 = 3

6 (2 x - 1) (3 x - 1) - 12 x^{2} = 7 x (2 x - 1)

x^{2} - 7 x - 14 = 0

What are the solutions to the equation 0=x^2+4x+16 ?
The solutions to the equation 0=x^2+4x+16 are x=-2+2sqrt(3)i,x=-2-2sqrt(3)i
Find the zeros of 0=x^2+4x+16
The zeros of 0=x^2+4x+16 are x=-2+2sqrt(3)i,x=-2-2sqrt(3)i