What are the solutions to the equation 33^2+b^2=34^2 ?

The solutions to the equation 33^2+b^2=34^2 are b=sqrt(67),b=-sqrt(67)

Find the zeros of 33^2+b^2=34^2

The zeros of 33^2+b^2=34^2 are b=sqrt(67),b=-sqrt(67)

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3 3^{2} + b^{2} = 3 4^{2}

b = 67, b = - 67

Decimal Notation

b = 8.18535 \dots, b = - 8.18535 \dots

Solution steps

3 3^{2} + b^{2} = 3 4^{2}

Move

3 3^{2}

to the right side

b^{2} = 67

For

x^{2} = f (a)

the solutions are

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

b = 67, b = - 67

- 84 = - 2 x + 5 (3 x + 4)

1.5 x + 4 (303 - x) = 942

4 ab - 2 b + 6 a - 3

2 x^{2} + 9 x

m \cdot (m + n) - 3 m \cdot (m + n)^{2}

What are the solutions to the equation 33^2+b^2=34^2 ?
The solutions to the equation 33^2+b^2=34^2 are b=sqrt(67),b=-sqrt(67)
Find the zeros of 33^2+b^2=34^2
The zeros of 33^2+b^2=34^2 are b=sqrt(67),b=-sqrt(67)