What are the solutions to the equation (x-3)^2+4=-x^2 ?

The solutions to the equation (x-3)^2+4=-x^2 are x= 3/2+i(sqrt(17))/2 ,x= 3/2-i(sqrt(17))/2

Find the zeros of (x-3)^2+4=-x^2

The zeros of (x-3)^2+4=-x^2 are x= 3/2+i(sqrt(17))/2 ,x= 3/2-i(sqrt(17))/2

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(x - 3)^{2} + 4 = - x^{2}

x = \frac{3}{2} + i \frac{17}{2}, x = \frac{3}{2} - i \frac{17}{2}

Solution steps

(x - 3)^{2} + 4 = - x^{2}

Expand

(x - 3)^{2} + 4 : x^{2} - 6 x + 13

x^{2} - 6 x + 13 = - x^{2}

Move

x^{2}

to the left side

2 x^{2} - 6 x + 13 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 2 \cdot 13}{2 \cdot 2}

Simplify

(- 6)^{2} - 4 \cdot 2 \cdot 13 : 217 i

x_{1, 2} = \frac{- ( - 6 ) \pm 2 17 i}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 6 ) + 2 17 i}{2 \cdot 2}, x_{2} = \frac{- ( - 6 ) - 2 17 i}{2 \cdot 2}

x = \frac{- ( - 6 ) + 2 17 i}{2 \cdot 2} : \frac{3}{2} + i \frac{17}{2}

x = \frac{- ( - 6 ) - 2 17 i}{2 \cdot 2} : \frac{3}{2} - i \frac{17}{2}

The solutions to the quadratic equation are:

x = \frac{3}{2} + i \frac{17}{2}, x = \frac{3}{2} - i \frac{17}{2}

2 x^{2} - 5 x = x^{2} + 2 x - 1

9 x^{2} - 12 x = 14

(x - 2) (2 x + 4) = 12

2 x^{2} - 1 = 11 x + 5

3 x^{2} - 12 x + 13

What are the solutions to the equation (x-3)^2+4=-x^2 ?
The solutions to the equation (x-3)^2+4=-x^2 are x= 3/2+i(sqrt(17))/2 ,x= 3/2-i(sqrt(17))/2
Find the zeros of (x-3)^2+4=-x^2
The zeros of (x-3)^2+4=-x^2 are x= 3/2+i(sqrt(17))/2 ,x= 3/2-i(sqrt(17))/2