What are the solutions to the equation (x+2)^2+(x-3)^2=9 ?

The solutions to the equation (x+2)^2+(x-3)^2=9 are x= 1/2+i(sqrt(7))/2 ,x= 1/2-i(sqrt(7))/2

Find the zeros of (x+2)^2+(x-3)^2=9

The zeros of (x+2)^2+(x-3)^2=9 are x= 1/2+i(sqrt(7))/2 ,x= 1/2-i(sqrt(7))/2

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(x + 2)^{2} + (x - 3)^{2} = 9

x = \frac{1}{2} + i \frac{7}{2}, x = \frac{1}{2} - i \frac{7}{2}

Solution steps

(x + 2)^{2} + (x - 3)^{2} = 9

Expand

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

2 x^{2} - 2 x + 13 = 9

Move

9

to the left side

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

Solve with the quadratic formula

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

Simplify

(- 2)^{2} - 4 \cdot 2 \cdot 4 : 27 i

x_{1, 2} = \frac{- ( - 2 ) \pm 2 7 i}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 2 7 i}{2 \cdot 2}, x_{2} = \frac{- ( - 2 ) - 2 7 i}{2 \cdot 2}

x = \frac{- ( - 2 ) + 2 7 i}{2 \cdot 2} : \frac{1}{2} + i \frac{7}{2}

x = \frac{- ( - 2 ) - 2 7 i}{2 \cdot 2} : \frac{1}{2} - i \frac{7}{2}

The solutions to the quadratic equation are:

x = \frac{1}{2} + i \frac{7}{2}, x = \frac{1}{2} - i \frac{7}{2}

- \frac{1}{6} x^{2} + x - 3 = 0

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

a^{2} - 1 = 0

3 x^{2} - 1 = 47

y^{2} + 5 y + 4 = 0

What are the solutions to the equation (x+2)^2+(x-3)^2=9 ?
The solutions to the equation (x+2)^2+(x-3)^2=9 are x= 1/2+i(sqrt(7))/2 ,x= 1/2-i(sqrt(7))/2
Find the zeros of (x+2)^2+(x-3)^2=9
The zeros of (x+2)^2+(x-3)^2=9 are x= 1/2+i(sqrt(7))/2 ,x= 1/2-i(sqrt(7))/2