What are the solutions to the equation b^2-4=0 ?

The solutions to the equation b^2-4=0 are b=2,b=-2

Find the zeros of b^2-4=0

The zeros of b^2-4=0 are b=2,b=-2

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b^{2} - 4 = 0

b = 2, b = - 2

Solution steps

b^{2} - 4 = 0

Move

4

to the right side

b^{2} = 4

For

x^{2} = f (a)

the solutions are

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

b = 4, b = - 4

4 = 2

- 4 = - 2

b = 2, b = - 2

x^{2} \cdot \frac{157}{8} + \frac{51}{4} \cdot \frac{4553}{816} = 0

\frac{x}{2} - \frac{x}{5} = 2

\frac{1}{2} (x + 2) + \frac{3}{4} (x + 4) = x + 5

\frac{5}{8} \cdot \frac{3}{5}

5 a + 18 < - 27

What are the solutions to the equation b^2-4=0 ?
The solutions to the equation b^2-4=0 are b=2,b=-2
Find the zeros of b^2-4=0
The zeros of b^2-4=0 are b=2,b=-2