What are the solutions to the equation (2x-1)^2+11=4 ?

The solutions to the equation (2x-1)^2+11=4 are x= 1/2+i(sqrt(7))/2 ,x= 1/2-i(sqrt(7))/2

Find the zeros of (2x-1)^2+11=4

The zeros of (2x-1)^2+11=4 are x= 1/2+i(sqrt(7))/2 ,x= 1/2-i(sqrt(7))/2

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(2 x - 1)^{2} + 11 = 4

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

Solution steps

(2 x - 1)^{2} + 11 = 4

Move

11

to the right side

(2 x - 1)^{2} = - 7

For

(g (x))^{2} = f (a)

the solutions are

g (x) = f (a), - f (a)

Solve

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

Solve

2 x - 1 = - - 7 : x = \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}

4 v^{2} + 9 v - 2 = 4 v

3 x^{2} + 17 x + 15 = 5

a, g (a + b) = (a + b)^{2} - 4 (a + b)

2 \cdot (x^{2} + 10) = 22

x, x^{2} - x^{2} y^{3} + 5 y^{4} = 0

What are the solutions to the equation (2x-1)^2+11=4 ?
The solutions to the equation (2x-1)^2+11=4 are x= 1/2+i(sqrt(7))/2 ,x= 1/2-i(sqrt(7))/2
Find the zeros of (2x-1)^2+11=4
The zeros of (2x-1)^2+11=4 are x= 1/2+i(sqrt(7))/2 ,x= 1/2-i(sqrt(7))/2