What are the solutions to the equation (14-b)^3+b^3=812 ?

The solutions to the equation (14-b)^3+b^3=812 are b=7+sqrt(3),b=7-sqrt(3)

Find the zeros of (14-b)^3+b^3=812

The zeros of (14-b)^3+b^3=812 are b=7+sqrt(3),b=7-sqrt(3)

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(14 - b)^{3} + b^{3} = 812

b = 7 + 3, b = 7 - 3

Decimal Notation

b = 8.73205 \dots, b = 5.26794 \dots

Solution steps

(14 - b)^{3} + b^{3} = 812

Expand

(14 - b)^{3} + b^{3} : 2744 - 588 b + 42 b^{2}

2744 - 588 b + 42 b^{2} = 812

Move

812

to the left side

42 b^{2} - 588 b + 1932 = 0

Solve with the quadratic formula

b_{1, 2} = \frac{- ( - 588 ) \pm ( - 588 ) ^{2} - 4 \cdot 42 \cdot 1932}{2 \cdot 42}

(- 588)^{2} - 4 \cdot 42 \cdot 1932 = 843

b_{1, 2} = \frac{- ( - 588 ) \pm 84 3}{2 \cdot 42}

Separate the solutions

b_{1} = \frac{- ( - 588 ) + 84 3}{2 \cdot 42}, b_{2} = \frac{- ( - 588 ) - 84 3}{2 \cdot 42}

b = \frac{- ( - 588 ) + 84 3}{2 \cdot 42} : 7 + 3

b = \frac{- ( - 588 ) - 84 3}{2 \cdot 42} : 7 - 3

The solutions to the quadratic equation are:

b = 7 + 3, b = 7 - 3

10 = π r^{2}

4.9 x^{2} + 5 x - 10 = 0

- 7 x^{2} + 2 x = - 9

4 + 9 n - 3 n^{2} = 2 - n

25 x^{2} = 0

What are the solutions to the equation (14-b)^3+b^3=812 ?
The solutions to the equation (14-b)^3+b^3=812 are b=7+sqrt(3),b=7-sqrt(3)
Find the zeros of (14-b)^3+b^3=812
The zeros of (14-b)^3+b^3=812 are b=7+sqrt(3),b=7-sqrt(3)