What are the solutions to the equation (6d+1)^2+12=13 ?

The solutions to the equation (6d+1)^2+12=13 are d=0,d=-1/3

Find the zeros of (6d+1)^2+12=13

The zeros of (6d+1)^2+12=13 are d=0,d=-1/3

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(6 d + 1)^{2} + 12 = 13

d = 0, d = - \frac{1}{3}

Decimal Notation

d = 0, d = - 0.33333 \dots

Solution steps

(6 d + 1)^{2} + 12 = 13

Move

12

to the right side

(6 d + 1)^{2} = 1

For

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

the solutions are

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

Solve

6 d + 1 = 1 : d = 0

Solve

6 d + 1 = - 1 : d = - \frac{1}{3}

The solutions to the quadratic equation are:

d = 0, d = - \frac{1}{3}

- \frac{2}{6} \cdot t^{3} + 2 t + 1

x y + 3 x - y - 3

2 x + 7 = 6 x + (- 1)

54 y^{5} - 2 y^{2}

3 x^{2} + 10 x y + 3 y^{2}

What are the solutions to the equation (6d+1)^2+12=13 ?
The solutions to the equation (6d+1)^2+12=13 are d=0,d=-1/3
Find the zeros of (6d+1)^2+12=13
The zeros of (6d+1)^2+12=13 are d=0,d=-1/3