What are the solutions to the equation (6x+2)^2+4=28 ?

The solutions to the equation (6x+2)^2+4=28 are x=(sqrt(6)-1)/3 ,x=-(1+sqrt(6))/3

Find the zeros of (6x+2)^2+4=28

The zeros of (6x+2)^2+4=28 are x=(sqrt(6)-1)/3 ,x=-(1+sqrt(6))/3

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(6 x + 2)^{2} + 4 = 28

x = \frac{6 - 1}{3}, x = - \frac{1 + 6}{3}

Solution steps

(6 x + 2)^{2} + 4 = 28

Move

4

to the right side

(6 x + 2)^{2} = 24

For

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

the solutions are

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

Solve

6 x + 2 = 24 : x = \frac{6 - 1}{3}

Solve

6 x + 2 = - 24 : x = - \frac{1 + 6}{3}

The solutions to the quadratic equation are:

x = \frac{6 - 1}{3}, x = - \frac{1 + 6}{3}

\frac{4 ^{n}}{2 ^{m}}

125 a^{3} - 64

80

\frac{8}{4 + 3 i}

x^{2} + 12 x + 20 \geq 0

What are the solutions to the equation (6x+2)^2+4=28 ?
The solutions to the equation (6x+2)^2+4=28 are x=(sqrt(6)-1)/3 ,x=-(1+sqrt(6))/3
Find the zeros of (6x+2)^2+4=28
The zeros of (6x+2)^2+4=28 are x=(sqrt(6)-1)/3 ,x=-(1+sqrt(6))/3