What are the solutions to the equation x^3+2x^2-6x-2=x^3+x^2-7x+18 ?

The solutions to the equation x^3+2x^2-6x-2=x^3+x^2-7x+18 are x=4,x=-5

Find the zeros of x^3+2x^2-6x-2=x^3+x^2-7x+18

The zeros of x^3+2x^2-6x-2=x^3+x^2-7x+18 are x=4,x=-5

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x^{3} + 2 x^{2} - 6 x - 2 = x^{3} + x^{2} - 7 x + 18

x = 4, x = - 5

Solution steps

x^{3} + 2 x^{2} - 6 x - 2 = x^{3} + x^{2} - 7 x + 18

Subtract

x^{3} + x^{2} - 7 x + 18

from both sides

x^{3} + 2 x^{2} - 6 x - 2 - (x^{3} + x^{2} - 7 x + 18) = x^{3} + x^{2} - 7 x + 18 - (x^{3} + x^{2} - 7 x + 18)

Simplify

x^{2} + x - 20 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 20 )}{2 \cdot 1}

1^{2} - 4 \cdot 1 \cdot (- 20) = 9

x_{1, 2} = \frac{- 1 \pm 9}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 1 + 9}{2 \cdot 1}, x_{2} = \frac{- 1 - 9}{2 \cdot 1}

x = \frac{- 1 + 9}{2 \cdot 1} : 4

x = \frac{- 1 - 9}{2 \cdot 1} : - 5

The solutions to the quadratic equation are:

x = 4, x = - 5

\frac{3 x}{5} = \frac{4 x}{3} - 22

p^{2} - 144 = 0

- 5 - \frac{1}{2}

\frac{0.41 - 0.4}{\frac{24}{n}} < - 1.96

2 y^{3} - 72 x^{2} y

What are the solutions to the equation x^3+2x^2-6x-2=x^3+x^2-7x+18 ?
The solutions to the equation x^3+2x^2-6x-2=x^3+x^2-7x+18 are x=4,x=-5
Find the zeros of x^3+2x^2-6x-2=x^3+x^2-7x+18
The zeros of x^3+2x^2-6x-2=x^3+x^2-7x+18 are x=4,x=-5