What are the solutions to the equation (2x-3)^2-(x-5)^2=-23 ?

The solutions to the equation (2x-3)^2-(x-5)^2=-23 are x= 1/3+i(2sqrt(5))/3 ,x= 1/3-i(2sqrt(5))/3

Find the zeros of (2x-3)^2-(x-5)^2=-23

The zeros of (2x-3)^2-(x-5)^2=-23 are x= 1/3+i(2sqrt(5))/3 ,x= 1/3-i(2sqrt(5))/3

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(2 x - 3)^{2} - (x - 5)^{2} = - 23

x = \frac{1}{3} + i \frac{2 5}{3}, x = \frac{1}{3} - i \frac{2 5}{3}

Solution steps

(2 x - 3)^{2} - (x - 5)^{2} = - 23

Expand

(2 x - 3)^{2} - (x - 5)^{2} : 3 x^{2} - 2 x - 16

3 x^{2} - 2 x - 16 = - 23

Move

23

to the left side

3 x^{2} - 2 x + 7 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 3 \cdot 7}{2 \cdot 3}

Simplify

(- 2)^{2} - 4 \cdot 3 \cdot 7 : 45 i

x_{1, 2} = \frac{- ( - 2 ) \pm 4 5 i}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 4 5 i}{2 \cdot 3}, x_{2} = \frac{- ( - 2 ) - 4 5 i}{2 \cdot 3}

x = \frac{- ( - 2 ) + 4 5 i}{2 \cdot 3} : \frac{1}{3} + i \frac{2 5}{3}

x = \frac{- ( - 2 ) - 4 5 i}{2 \cdot 3} : \frac{1}{3} - i \frac{2 5}{3}

The solutions to the quadratic equation are:

x = \frac{1}{3} + i \frac{2 5}{3}, x = \frac{1}{3} - i \frac{2 5}{3}

30 x^{2} - 60 = 0

x^{2} + 11 x = 60

0 = x^{2} - x + 6

x^{2} - 6 x = 15

7 = 3 x^{2} - 18 x + 7

What are the solutions to the equation (2x-3)^2-(x-5)^2=-23 ?
The solutions to the equation (2x-3)^2-(x-5)^2=-23 are x= 1/3+i(2sqrt(5))/3 ,x= 1/3-i(2sqrt(5))/3
Find the zeros of (2x-3)^2-(x-5)^2=-23
The zeros of (2x-3)^2-(x-5)^2=-23 are x= 1/3+i(2sqrt(5))/3 ,x= 1/3-i(2sqrt(5))/3