What are the solutions to the equation (2x+4)^2=21 ?

The solutions to the equation (2x+4)^2=21 are x=(sqrt(21)-4)/2 ,x=(-sqrt(21)-4)/2

Find the zeros of (2x+4)^2=21

The zeros of (2x+4)^2=21 are x=(sqrt(21)-4)/2 ,x=(-sqrt(21)-4)/2

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(2 x + 4)^{2} = 21

x = \frac{21 - 4}{2}, x = \frac{- 21 - 4}{2}

Solution steps

(2 x + 4)^{2} = 21

For

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

the solutions are

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

Solve

2 x + 4 = 21 : x = \frac{21 - 4}{2}

Solve

2 x + 4 = - 21 : x = \frac{- 21 - 4}{2}

The solutions to the quadratic equation are:

x = \frac{21 - 4}{2}, x = \frac{- 21 - 4}{2}

5 = - 0.04 t^{2} + 0.8 t + 10

x, z^{2} = 9 x^{2} + y^{2}

y, 3 y^{2} + 2 a y = - b

ab x^{2} + ab = a^{2} x + b^{2} x

x^{2} + 6 x = - 8

What are the solutions to the equation (2x+4)^2=21 ?
The solutions to the equation (2x+4)^2=21 are x=(sqrt(21)-4)/2 ,x=(-sqrt(21)-4)/2
Find the zeros of (2x+4)^2=21
The zeros of (2x+4)^2=21 are x=(sqrt(21)-4)/2 ,x=(-sqrt(21)-4)/2