What are the solutions to the equation (3x+2)^2+4=(x+1)^2 ?

The solutions to the equation (3x+2)^2+4=(x+1)^2 are x=-5/8+i(sqrt(31))/8 ,x=-5/8-i(sqrt(31))/8

Find the zeros of (3x+2)^2+4=(x+1)^2

The zeros of (3x+2)^2+4=(x+1)^2 are x=-5/8+i(sqrt(31))/8 ,x=-5/8-i(sqrt(31))/8

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(3 x + 2)^{2} + 4 = (x + 1)^{2}

x = - \frac{5}{8} + i \frac{31}{8}, x = - \frac{5}{8} - i \frac{31}{8}

Solution steps

(3 x + 2)^{2} + 4 = (x + 1)^{2}

Expand

(3 x + 2)^{2} + 4 : 9 x^{2} + 12 x + 8

Expand

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

9 x^{2} + 12 x + 8 = x^{2} + 2 x + 1

Move

1

to the left side

9 x^{2} + 12 x + 7 = x^{2} + 2 x

Move

2 x

to the left side

9 x^{2} + 10 x + 7 = x^{2}

Move

x^{2}

to the left side

8 x^{2} + 10 x + 7 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 8 \cdot 7}{2 \cdot 8}

Simplify

1 0^{2} - 4 \cdot 8 \cdot 7 : 231 i

x_{1, 2} = \frac{- 10 \pm 2 31 i}{2 \cdot 8}

Separate the solutions

x_{1} = \frac{- 10 + 2 31 i}{2 \cdot 8}, x_{2} = \frac{- 10 - 2 31 i}{2 \cdot 8}

x = \frac{- 10 + 2 31 i}{2 \cdot 8} : - \frac{5}{8} + i \frac{31}{8}

x = \frac{- 10 - 2 31 i}{2 \cdot 8} : - \frac{5}{8} - i \frac{31}{8}

The solutions to the quadratic equation are:

x = - \frac{5}{8} + i \frac{31}{8}, x = - \frac{5}{8} - i \frac{31}{8}

x, x^{2} - y^{2} + z^{2} = 0

t, f = (t - 5) e t - 5 U (t - 5)

2 (x + 5) = x^{2} + 5 x

v^{2} - 3 v - 18 = 0

t, u = t^{2} + 2

What are the solutions to the equation (3x+2)^2+4=(x+1)^2 ?
The solutions to the equation (3x+2)^2+4=(x+1)^2 are x=-5/8+i(sqrt(31))/8 ,x=-5/8-i(sqrt(31))/8
Find the zeros of (3x+2)^2+4=(x+1)^2
The zeros of (3x+2)^2+4=(x+1)^2 are x=-5/8+i(sqrt(31))/8 ,x=-5/8-i(sqrt(31))/8