What are the solutions to the equation x^3+4x^2-20x=80 ?

The solutions to the equation x^3+4x^2-20x=80 are x=-4,x=-2sqrt(5),x=2sqrt(5)

Find the zeros of x^3+4x^2-20x=80

The zeros of x^3+4x^2-20x=80 are x=-4,x=-2sqrt(5),x=2sqrt(5)

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x^{3} + 4 x^{2} - 20 x = 80

x = - 4, x = - 25, x = 25

Decimal Notation

x = - 4, x = - 4.47213 \dots, x = 4.47213 \dots

Solution steps

x^{3} + 4 x^{2} - 20 x = 80

Move

80

to the left side

x^{3} + 4 x^{2} - 20 x - 80 = 0

Factor

x^{3} + 4 x^{2} - 20 x - 80 : (x + 4) (x + 25) (x - 25)

(x + 4) (x + 25) (x - 25) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x + 4 = 0 or x + 25 = 0 or x - 25 = 0

Solve

x + 4 = 0 : x = - 4

Solve

x + 25 = 0 : x = - 25

Solve

x - 25 = 0 : x = 25

The solutions are

x = - 4, x = - 25, x = 25

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

x^{5} + 64 x^{2} = 0

x, 5 x^{3} - 8 = 2 y^{2}

x (3 + x)^{4} + (3 - x)^{4} = 386

t^{6} - t^{2} = 0

What are the solutions to the equation x^3+4x^2-20x=80 ?
The solutions to the equation x^3+4x^2-20x=80 are x=-4,x=-2sqrt(5),x=2sqrt(5)
Find the zeros of x^3+4x^2-20x=80
The zeros of x^3+4x^2-20x=80 are x=-4,x=-2sqrt(5),x=2sqrt(5)