What are the solutions to the equation solvefor X,I=X^2+2X+4 ?

The solutions to the equation solvefor X,I=X^2+2X+4 are X=-1+sqrt(I-3),X=-sqrt(I-3)-1

Find the zeros of solvefor X,I=X^2+2X+4

The zeros of solvefor X,I=X^2+2X+4 are X=-1+sqrt(I-3),X=-sqrt(I-3)-1

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solve for

X, I = X^{2} + 2 X + 4

X = - 1 + I - 3, X = - I - 3 - 1

Solution steps

I = X^{2} + 2 X + 4

Switch sides

X^{2} + 2 X + 4 = I

Move

I

to the left side

X^{2} + 2 X + 4 - I = 0

Solve with the quadratic formula

X_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot ( 4 - I )}{2 \cdot 1}

Simplify

2^{2} - 4 \cdot 1 \cdot (4 - I) : 2 I - 3

X_{1, 2} = \frac{- 2 \pm 2 I - 3}{2 \cdot 1}

Separate the solutions

X_{1} = \frac{- 2 + 2 I - 3}{2 \cdot 1}, X_{2} = \frac{- 2 - 2 I - 3}{2 \cdot 1}

X = \frac{- 2 + 2 I - 3}{2 \cdot 1} : - 1 + I - 3

X = \frac{- 2 - 2 I - 3}{2 \cdot 1} : - I - 3 - 1

The solutions to the quadratic equation are:

X = - 1 + I - 3, X = - I - 3 - 1

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

2 (x - 3) (x + 1) = x (x - 1) + 2

6 x^{2} - 26 x + 24 = 0

8 x^{2} + 4 x - 3 = 0

- 4 x^{2} - 6 x + 54 = 0

What are the solutions to the equation solvefor X,I=X^2+2X+4 ?
The solutions to the equation solvefor X,I=X^2+2X+4 are X=-1+sqrt(I-3),X=-sqrt(I-3)-1
Find the zeros of solvefor X,I=X^2+2X+4
The zeros of solvefor X,I=X^2+2X+4 are X=-1+sqrt(I-3),X=-sqrt(I-3)-1