What are the solutions to the equation (2x+3)^2=20 ?

The solutions to the equation (2x+3)^2=20 are x=(2sqrt(5)-3)/2 ,x=(-2sqrt(5)-3)/2

Find the zeros of (2x+3)^2=20

The zeros of (2x+3)^2=20 are x=(2sqrt(5)-3)/2 ,x=(-2sqrt(5)-3)/2

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(2 x + 3)^{2} = 20

x = \frac{2 5 - 3}{2}, x = \frac{- 2 5 - 3}{2}

Decimal Notation

x = 0.73606 \dots, x = - 3.73606 \dots

Solution steps

(2 x + 3)^{2} = 20

For

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

the solutions are

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

Solve

2 x + 3 = 20 : x = \frac{2 5 - 3}{2}

Solve

2 x + 3 = - 20 : x = \frac{- 2 5 - 3}{2}

The solutions to the quadratic equation are:

x = \frac{2 5 - 3}{2}, x = \frac{- 2 5 - 3}{2}

x^{2} + 12 x + 8 = 0

2 x^{2} - 15 = 0

- 2 x^{2} + 3 x + 1 = 0

(x - 2)^{2} + 46 = 47

0.02 w^{2} - 0.3 w = 20

What are the solutions to the equation (2x+3)^2=20 ?
The solutions to the equation (2x+3)^2=20 are x=(2sqrt(5)-3)/2 ,x=(-2sqrt(5)-3)/2
Find the zeros of (2x+3)^2=20
The zeros of (2x+3)^2=20 are x=(2sqrt(5)-3)/2 ,x=(-2sqrt(5)-3)/2