What are the solutions to the equation 2(2x+1)^2=(x+3)^2-10 ?

The solutions to the equation 2(2x+1)^2=(x+3)^2-10 are x=-1/7+i(2sqrt(5))/7 ,x=-1/7-i(2sqrt(5))/7

Find the zeros of 2(2x+1)^2=(x+3)^2-10

The zeros of 2(2x+1)^2=(x+3)^2-10 are x=-1/7+i(2sqrt(5))/7 ,x=-1/7-i(2sqrt(5))/7

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2 (2 x + 1)^{2} = (x + 3)^{2} - 10

x = - \frac{1}{7} + i \frac{2 5}{7}, x = - \frac{1}{7} - i \frac{2 5}{7}

Solution steps

2 (2 x + 1)^{2} = (x + 3)^{2} - 10

Expand

2 (2 x + 1)^{2} : 8 x^{2} + 8 x + 2

Expand

(x + 3)^{2} - 10 : x^{2} + 6 x - 1

8 x^{2} + 8 x + 2 = x^{2} + 6 x - 1

Move

1

to the left side

8 x^{2} + 8 x + 3 = x^{2} + 6 x

Move

6 x

to the left side

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

Move

x^{2}

to the left side

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

Solve with the quadratic formula

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

Simplify

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

x_{1, 2} = \frac{- 2 \pm 4 5 i}{2 \cdot 7}

Separate the solutions

x_{1} = \frac{- 2 + 4 5 i}{2 \cdot 7}, x_{2} = \frac{- 2 - 4 5 i}{2 \cdot 7}

x = \frac{- 2 + 4 5 i}{2 \cdot 7} : - \frac{1}{7} + i \frac{2 5}{7}

x = \frac{- 2 - 4 5 i}{2 \cdot 7} : - \frac{1}{7} - i \frac{2 5}{7}

The solutions to the quadratic equation are:

x = - \frac{1}{7} + i \frac{2 5}{7}, x = - \frac{1}{7} - i \frac{2 5}{7}

x^{2} - 2 x = 3

7 x^{2} - 18 = 45

9 x^{2} - 6 x + 7 = 0

- 0.5 x^{2} + 2 x + 1.5 = 0

x^{2} + 5 x + 9 = 0

What are the solutions to the equation 2(2x+1)^2=(x+3)^2-10 ?
The solutions to the equation 2(2x+1)^2=(x+3)^2-10 are x=-1/7+i(2sqrt(5))/7 ,x=-1/7-i(2sqrt(5))/7
Find the zeros of 2(2x+1)^2=(x+3)^2-10
The zeros of 2(2x+1)^2=(x+3)^2-10 are x=-1/7+i(2sqrt(5))/7 ,x=-1/7-i(2sqrt(5))/7