What are the solutions to the equation 3/2 b^2-7=41 ?

The solutions to the equation 3/2 b^2-7=41 are b=4sqrt(2),b=-4sqrt(2)

Find the zeros of 3/2 b^2-7=41

The zeros of 3/2 b^2-7=41 are b=4sqrt(2),b=-4sqrt(2)

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\frac{3}{2} b^{2} - 7 = 41

b = 42, b = - 42

Decimal

b = 5.65685 \dots, b = - 5.65685 \dots

Solution steps

\frac{3}{2} b^{2} - 7 = 41

Move

7

to the right side

\frac{3}{2} b^{2} = 48

Multiply both sides by

2

3 b^{2} = 96

Divide both sides by

3

b^{2} = 32

For

x^{2} = f (a)

the solutions are

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

b = 32, b = - 32

32 = 42

- 32 = - 42

b = 42, b = - 42

12 = x^{2} - 2 x + 1

3 (x - 2 + 4) (2 x - 1) = 5 (x + 5) - 2 (x - 3)

0 = \frac{27 - x ^{2}}{2}

y^{2} + y + 1

x, x^{2} - 8 x = - x

What are the solutions to the equation 3/2 b^2-7=41 ?
The solutions to the equation 3/2 b^2-7=41 are b=4sqrt(2),b=-4sqrt(2)
Find the zeros of 3/2 b^2-7=41
The zeros of 3/2 b^2-7=41 are b=4sqrt(2),b=-4sqrt(2)