What are the solutions to the equation 4(n-2)4(n-2)=12 ?

The solutions to the equation 4(n-2)4(n-2)=12 are n=(sqrt(3))/2+2,n=-(sqrt(3))/2+2

Find the zeros of 4(n-2)4(n-2)=12

The zeros of 4(n-2)4(n-2)=12 are n=(sqrt(3))/2+2,n=-(sqrt(3))/2+2

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4 (n - 2) 4 (n - 2) = 12

n = \frac{3}{2} + 2, n = - \frac{3}{2} + 2

Decimal Notation

n = 2.86602 \dots, n = 1.13397 \dots

Solution steps

4 (n - 2) \cdot 4 (n - 2) = 12

Simplify

16 (n - 2)^{2} = 12

Divide both sides by

16

\frac{16 ( n - 2 ) ^{2}}{16} = \frac{12}{16}

(n - 2)^{2} = \frac{3}{4}

For

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

the solutions are

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

Solve

n - 2 = \frac{3}{4} : n = \frac{3}{2} + 2

Solve

n - 2 = - \frac{3}{4} : n = - \frac{3}{2} + 2

The solutions to the quadratic equation are:

n = \frac{3}{2} + 2, n = - \frac{3}{2} + 2

7.15 = (0.75 + 2 c) (0.3 + 2 c)

x^{2} = (17 mod \cdot 29)

- x^{2} + 2 x + 3 = x^{2} - 2 x + 3

x^{2} + 3 x = 26

\frac{25}{100} \cdot x + \frac{40}{100} (x \cdot x) = 178

What are the solutions to the equation 4(n-2)4(n-2)=12 ?
The solutions to the equation 4(n-2)4(n-2)=12 are n=(sqrt(3))/2+2,n=-(sqrt(3))/2+2
Find the zeros of 4(n-2)4(n-2)=12
The zeros of 4(n-2)4(n-2)=12 are n=(sqrt(3))/2+2,n=-(sqrt(3))/2+2