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

The solutions to the equation 2x^3+x^2-6x-7=2x^2+3x-7 are x=0,x=(1+sqrt(73))/4 ,x=(1-sqrt(73))/4

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

The zeros of 2x^3+x^2-6x-7=2x^2+3x-7 are x=0,x=(1+sqrt(73))/4 ,x=(1-sqrt(73))/4

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

x = 0, x = \frac{1 + 73}{4}, x = \frac{1 - 73}{4}

Solution steps

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

Move

7

to the right side

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

Move

3 x

to the left side

2 x^{3} + x^{2} - 9 x = 2 x^{2}

Move

2 x^{2}

to the left side

2 x^{3} - x^{2} - 9 x = 0

Factor

2 x^{3} - x^{2} - 9 x : x (2 x^{2} - x - 9)

x (2 x^{2} - x - 9) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x = 0 or 2 x^{2} - x - 9 = 0

Solve

2 x^{2} - x - 9 = 0 : x = \frac{1 + 73}{4}, x = \frac{1 - 73}{4}

The solutions are

x = 0, x = \frac{1 + 73}{4}, x = \frac{1 - 73}{4}

4 x^{3} = 756

47.24 = 2370.903 (b^{2008})

x^{4} + 9 x^{3} + x^{2} + 9 = 0

x^{3} - 7 x + 10 = 0

\frac{1}{2} t^{2} (t^{2} - 2 t + 1) = 0

What are the solutions to the equation 2x^3+x^2-6x-7=2x^2+3x-7 ?
The solutions to the equation 2x^3+x^2-6x-7=2x^2+3x-7 are x=0,x=(1+sqrt(73))/4 ,x=(1-sqrt(73))/4
Find the zeros of 2x^3+x^2-6x-7=2x^2+3x-7
The zeros of 2x^3+x^2-6x-7=2x^2+3x-7 are x=0,x=(1+sqrt(73))/4 ,x=(1-sqrt(73))/4