What are the solutions to the equation (3x+12)^2=196 ?

The solutions to the equation (3x+12)^2=196 are x= 2/3 ,x=-26/3

Find the zeros of (3x+12)^2=196

The zeros of (3x+12)^2=196 are x= 2/3 ,x=-26/3

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(3 x + 12)^{2} = 196

x = \frac{2}{3}, x = - \frac{26}{3}

Decimal Notation

x = 0.66666 \dots, x = - 8.66666 \dots

Solution steps

(3 x + 12)^{2} = 196

For

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

the solutions are

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

Solve

3 x + 12 = 196 : x = \frac{2}{3}

Solve

3 x + 12 = - 196 : x = - \frac{26}{3}

The solutions to the quadratic equation are:

x = \frac{2}{3}, x = - \frac{26}{3}

- 4 x^{2} + 84 = 4

- X^{2} + 7 X - 10 = 0

x, - 12 x^{2} + 16 x = - 192

t^{2} + 20 t = 0

p^{2} - 3 p + 5 = 0

What are the solutions to the equation (3x+12)^2=196 ?
The solutions to the equation (3x+12)^2=196 are x= 2/3 ,x=-26/3
Find the zeros of (3x+12)^2=196
The zeros of (3x+12)^2=196 are x= 2/3 ,x=-26/3