What are the solutions to the equation 3x^3-5x^2+2=0 ?

The solutions to the equation 3x^3-5x^2+2=0 are x=1,x=(1+sqrt(7))/3 ,x=(1-sqrt(7))/3

Find the zeros of 3x^3-5x^2+2=0

The zeros of 3x^3-5x^2+2=0 are x=1,x=(1+sqrt(7))/3 ,x=(1-sqrt(7))/3

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3 x^{3} - 5 x^{2} + 2 = 0

x = 1, x = \frac{1 + 7}{3}, x = \frac{1 - 7}{3}

Decimal Notation

x = 1, x = 1.21525 \dots, x = - 0.54858 \dots

Solution steps

3 x^{3} - 5 x^{2} + 2 = 0

Factor

3 x^{3} - 5 x^{2} + 2 : (x - 1) (3 x^{2} - 2 x - 2)

(x - 1) (3 x^{2} - 2 x - 2) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 1 = 0 or 3 x^{2} - 2 x - 2 = 0

Solve

x - 1 = 0 : x = 1

Solve

3 x^{2} - 2 x - 2 = 0 : x = \frac{1 + 7}{3}, x = \frac{1 - 7}{3}

The solutions are

x = 1, x = \frac{1 + 7}{3}, x = \frac{1 - 7}{3}

x (x^{2} + 4) = 0

m^{3} + 4 m = 0

r^{4} + r^{3} - 7 r^{2} - r + 6 = 0

4 x - 4 x^{9} = 0

y, (x^{2} + y^{2})^{2} = x^{2} - y^{2}

What are the solutions to the equation 3x^3-5x^2+2=0 ?
The solutions to the equation 3x^3-5x^2+2=0 are x=1,x=(1+sqrt(7))/3 ,x=(1-sqrt(7))/3
Find the zeros of 3x^3-5x^2+2=0
The zeros of 3x^3-5x^2+2=0 are x=1,x=(1+sqrt(7))/3 ,x=(1-sqrt(7))/3