What are the solutions to the equation 5(n+1)^2=40 ?

The solutions to the equation 5(n+1)^2=40 are n=2sqrt(2)-1,n=-2sqrt(2)-1

Find the zeros of 5(n+1)^2=40

The zeros of 5(n+1)^2=40 are n=2sqrt(2)-1,n=-2sqrt(2)-1

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5 (n + 1)^{2} = 40

n = 22 - 1, n = - 22 - 1

Decimal Notation

n = 1.82842 \dots, n = - 3.82842 \dots

Solution steps

5 (n + 1)^{2} = 40

Divide both sides by

5

(n + 1)^{2} = 8

For

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

the solutions are

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

Solve

n + 1 = 8 : n = 22 - 1

Solve

n + 1 = - 8 : n = - 22 - 1

The solutions to the quadratic equation are:

n = 22 - 1, n = - 22 - 1

(\frac{- 2 x ^{- 2} y ^{4}}{x ^{2} y ^{6}})^{3}

(3 x - 2 y)^{2} - (2 x - 3 y)^{2}

6 + 2 n = 3

119 - 0.16 \cdot 28 \cdot 9.8 = 28 \cdot x

\frac{1}{2} \cdot 12 \cdot 5

What are the solutions to the equation 5(n+1)^2=40 ?
The solutions to the equation 5(n+1)^2=40 are n=2sqrt(2)-1,n=-2sqrt(2)-1
Find the zeros of 5(n+1)^2=40
The zeros of 5(n+1)^2=40 are n=2sqrt(2)-1,n=-2sqrt(2)-1