What are the solutions to the equation 5b^2-4=41 ?

The solutions to the equation 5b^2-4=41 are b=3,b=-3

Find the zeros of 5b^2-4=41

The zeros of 5b^2-4=41 are b=3,b=-3

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5 b^{2} - 4 = 41

b = 3, b = - 3

Solution steps

5 b^{2} - 4 = 41

Move

4

to the right side

5 b^{2} = 45

Divide both sides by

5

b^{2} = 9

For

x^{2} = f (a)

the solutions are

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

b = 9, b = - 9

9 = 3

- 9 = - 3

b = 3, b = - 3

2 t^{2} - 5 t - 3 = 0

4 2 x - 3 < 12

\frac{1}{9} \cdot \frac{5}{4}

\frac{- 6 \cdot 1 0 ^{- 9}}{8.85 \cdot 1 0 ^{- 12}}

- 34

What are the solutions to the equation 5b^2-4=41 ?
The solutions to the equation 5b^2-4=41 are b=3,b=-3
Find the zeros of 5b^2-4=41
The zeros of 5b^2-4=41 are b=3,b=-3