What are the solutions to the equation (2x-3)^2+7=0 ?

The solutions to the equation (2x-3)^2+7=0 are x= 3/2+i(sqrt(7))/2 ,x= 3/2-i(sqrt(7))/2

Find the zeros of (2x-3)^2+7=0

The zeros of (2x-3)^2+7=0 are x= 3/2+i(sqrt(7))/2 ,x= 3/2-i(sqrt(7))/2

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

x = \frac{3}{2} + i \frac{7}{2}, x = \frac{3}{2} - i \frac{7}{2}

Solution steps

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

Move

7

to the right side

(2 x - 3)^{2} = - 7

For

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

the solutions are

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

Solve

2 x - 3 = - 7 : x = \frac{3}{2} + i \frac{7}{2}

Solve

2 x - 3 = - - 7 : x = \frac{3}{2} - i \frac{7}{2}

The solutions to the quadratic equation are:

x = \frac{3}{2} + i \frac{7}{2}, x = \frac{3}{2} - i \frac{7}{2}

x^{2} - 1 = - 3

10 x^{2} - 5 x - 30 = 0

0 = 100 + 87 t - 5 t^{2}

x^{2} + 59 = 16 x

\frac{( 3 c ^{2} )}{2} = 3 (98)

What are the solutions to the equation (2x-3)^2+7=0 ?
The solutions to the equation (2x-3)^2+7=0 are x= 3/2+i(sqrt(7))/2 ,x= 3/2-i(sqrt(7))/2
Find the zeros of (2x-3)^2+7=0
The zeros of (2x-3)^2+7=0 are x= 3/2+i(sqrt(7))/2 ,x= 3/2-i(sqrt(7))/2