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

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

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

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

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

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

Decimal Notation

x = - 0.17712 \dots, x = - 2.82287 \dots

Solution steps

(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{7 - 3}{2}

Solve

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

The solutions to the quadratic equation are:

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

4 x (x + 5) = 0

x, 5 x^{2} + x y^{2} = c

36 d^{2} + 60 d + 25 = 16

112 x^{2} - 56 x + 3 = 0

(k - 1) y^{2} - 5 y + 3 k - 7 = 0

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