What are the solutions to the equation -x^3+(x-5)^3+1115=0 ?

The solutions to the equation -x^3+(x-5)^3+1115=0 are x=-6,x=11

Find the zeros of -x^3+(x-5)^3+1115=0

The zeros of -x^3+(x-5)^3+1115=0 are x=-6,x=11

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- x^{3} + (x - 5)^{3} + 1115 = 0

x = - 6, x = 11

Solution steps

- x^{3} + (x - 5)^{3} + 1115 = 0

Expand

- x^{3} + (x - 5)^{3} + 1115 : - 15 x^{2} + 75 x + 990

- 15 x^{2} + 75 x + 990 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 75 \pm 7 5 ^{2} - 4 ( - 15 ) \cdot 990}{2 ( - 15 )}

7 5^{2} - 4 (- 15) \cdot 990 = 255

x_{1, 2} = \frac{- 75 \pm 255}{2 ( - 15 )}

Separate the solutions

x_{1} = \frac{- 75 + 255}{2 ( - 15 )}, x_{2} = \frac{- 75 - 255}{2 ( - 15 )}

x = \frac{- 75 + 255}{2 ( - 15 )} : - 6

x = \frac{- 75 - 255}{2 ( - 15 )} : 11

The solutions to the quadratic equation are:

x = - 6, x = 11

\frac{2}{7} \cdot x^{2} + \frac{3}{4} \cdot x = \frac{1}{4} \cdot x

x, \frac{x ^{2}}{9} - \frac{y ^{2}}{16} = \frac{z}{4}

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

\frac{x ^{2} - ( 17 x )}{4 + 1} = 0

2 p^{2} + k^{3} = 15

What are the solutions to the equation -x^3+(x-5)^3+1115=0 ?
The solutions to the equation -x^3+(x-5)^3+1115=0 are x=-6,x=11
Find the zeros of -x^3+(x-5)^3+1115=0
The zeros of -x^3+(x-5)^3+1115=0 are x=-6,x=11