2 = 3x - 5x^3?

2 = 3x - 5x³, or

5x³ - 3x + 2 = 0

(x + 1)(5x² - 5x + 2) = 0

If the product of two factors equals zero,

then one or both factors equal zero.

If x + 1 = 0,

x = - 1

If 5x² - 5x + 2 = 0,

5x² - 5x = - 2

5(x² - x) = - 2

x² - x = - 2/5

x² - x = - 0.4

Completing the square,

x² - x + 0.25 = 0.25 - 0.4

(x - 0.5)² = - 0.15

x - 0.5 = √(- 0.15)

x - 0.5 = √[(0.15)(- 1)]

x - 0.5 = √(0.15) √(- 1)

x - 0.5 = ± 0.3873 i

x = 0.5 ± 0.3873 i

Since these roots are imaginary,

x = - 1

¯¯¯¯¯

2 = 3x − 5x³

=> 5x³ − 3x + 2 = 0

using the possible rational roots test find factors of the coefficient of the x⁰ term [2 = {±1, ±2}] and divide them by the factors of the coefficient of the xⁿ term [5 = {±1, ±5}] so that possible rational roots are {±1/5, ±2/5, ±1, ± 2}

now test each possible root in the equation until you find one that works

=> 5(-1)³ − 3(-1) + 2 = 0

so (x + 1) is a factor of 5x³ − 3x + 2

now use polynomial long division to factor out that factor

x + 1 ) 5x³ + 0x² − 3x + 2

5x²(x + 1) = 5x³ + 5x²

5x³ + 0x² − (5x³ + 5x²) = -5x²

-5x(x + 1) = -5x² − 5x

-5x² − 3x − (-5x² − 5x) = 2x

2(x + 1) = 2x + 2

2x + 2 − (2x + 2) = 0

=> (x + 1)(5x² − 5x + 2) = 0

now we can use our normal quadratic methods on 5x² − 5x + 2 = 0

=> 5x² − 5x = -2

=> x² − x = -2/5

=> x² − x + (-1/2)² = (-1/2)² − 2/5

=> (x − 1/2)² = 1/4 − 2/5

=> (x − 1/2)² = 5/20 − 8/20

=> (x − 1/2)² = -3/20

=> x − 1/2 = ±i√(3)/2√5

=> x − 1/2 = ±i√(3)/2√5 * √(5)/√5

=> x − 1/2 = ±i√(15)/10

=> x = 1/2 ±i√(15)/10

=> x = 5/10 ±i√(15)/10

=> x = (5 ± i√15)/10

so the values of x are x = -1, (5 − i√15)/10 and (5 + i√15)/10