What are the solutions to the equation (2(m+1)^2)/3 =9 ?

The solutions to the equation (2(m+1)^2)/3 =9 are m=(3sqrt(3))/(sqrt(2))-1,m=-(3sqrt(3))/(sqrt(2))-1

Find the zeros of (2(m+1)^2)/3 =9

The zeros of (2(m+1)^2)/3 =9 are m=(3sqrt(3))/(sqrt(2))-1,m=-(3sqrt(3))/(sqrt(2))-1

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\frac{2 ( m + 1 ) ^{2}}{3} = 9

m = \frac{3 3}{2} - 1, m = - \frac{3 3}{2} - 1

Decimal Notation

m = 2.67423 \dots, m = - 4.67423 \dots

Solution steps

\frac{2 ( m + 1 ) ^{2}}{3} = 9

Multiply both sides by

3

2 (m + 1)^{2} = 27

Divide both sides by

2

(m + 1)^{2} = \frac{27}{2}

For

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

the solutions are

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

Solve

m + 1 = \frac{27}{2} : m = \frac{3 3}{2} - 1

Solve

m + 1 = - \frac{27}{2} : m = - \frac{3 3}{2} - 1

The solutions to the quadratic equation are:

m = \frac{3 3}{2} - 1, m = - \frac{3 3}{2} - 1

x, y = a^{l} o g x^{2}

x, (x^{2} + y^{2} - 1) 3 - x^{2} y^{2} = 0

a^{2} \cdot (x - b) - b^{2} \cdot (x - a) = x^{2} \cdot (a - b)

2 x - 4 - x^{2} + 4 = 0

a, a^{2} + b^{2} + c^{2} = a + c - \frac{1}{2}

What are the solutions to the equation (2(m+1)^2)/3 =9 ?
The solutions to the equation (2(m+1)^2)/3 =9 are m=(3sqrt(3))/(sqrt(2))-1,m=-(3sqrt(3))/(sqrt(2))-1
Find the zeros of (2(m+1)^2)/3 =9
The zeros of (2(m+1)^2)/3 =9 are m=(3sqrt(3))/(sqrt(2))-1,m=-(3sqrt(3))/(sqrt(2))-1