What are the solutions to the equation 3(x+2)^2+4=40 ?

The solutions to the equation 3(x+2)^2+4=40 are x=2sqrt(3)-2,x=-2sqrt(3)-2

Find the zeros of 3(x+2)^2+4=40

The zeros of 3(x+2)^2+4=40 are x=2sqrt(3)-2,x=-2sqrt(3)-2

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3 (x + 2)^{2} + 4 = 40

x = 23 - 2, x = - 23 - 2

Decimal Notation

x = 1.46410 \dots, x = - 5.46410 \dots

Solution steps

3 (x + 2)^{2} + 4 = 40

Move

4

to the right side

3 (x + 2)^{2} = 36

Divide both sides by

3

(x + 2)^{2} = 12

For

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

the solutions are

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

Solve

x + 2 = 12 : x = 23 - 2

Solve

x + 2 = - 12 : x = - 23 - 2

The solutions to the quadratic equation are:

x = 23 - 2, x = - 23 - 2

3 x (2 x + 9) = 10 x

4 x^{2} - 11 x = 7

7 c^{2} = c + 6

4 c^{2} + c - 6 = 0

9 x + 28 = 9 x^{2}

What are the solutions to the equation 3(x+2)^2+4=40 ?
The solutions to the equation 3(x+2)^2+4=40 are x=2sqrt(3)-2,x=-2sqrt(3)-2
Find the zeros of 3(x+2)^2+4=40
The zeros of 3(x+2)^2+4=40 are x=2sqrt(3)-2,x=-2sqrt(3)-2