What are the solutions to the equation 20=x^3+4x^2-5x ?

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

Find the zeros of 20=x^3+4x^2-5x

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

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20 = x^{3} + 4 x^{2} - 5 x

x = - 4, x = - 5, x = 5

Decimal Notation

x = - 4, x = - 2.23606 \dots, x = 2.23606 \dots

Solution steps

20 = x^{3} + 4 x^{2} - 5 x

Switch sides

x^{3} + 4 x^{2} - 5 x = 20

Move

20

to the left side

x^{3} + 4 x^{2} - 5 x - 20 = 0

Factor

x^{3} + 4 x^{2} - 5 x - 20 : (x + 4) (x + 5) (x - 5)

(x + 4) (x + 5) (x - 5) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x + 4 = 0 or x + 5 = 0 or x - 5 = 0

Solve

x + 4 = 0 : x = - 4

Solve

x + 5 = 0 : x = - 5

Solve

x - 5 = 0 : x = 5

The solutions are

x = - 4, x = - 5, x = 5

x^{3} - 343 = 0

(x - 1)^{3} + 8 = 0

8 x^{3} - 14 x^{2} + 7 x - 1 = 0

x^{4} - 10 x^{2} = - 9

z^{3} + 64 = 0

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