(3x^3+4x^2) -(2x^3+3x^2-x)=(3x 3 +4x 2 )−(2x 3 +3x 2 −x)

x = 0 or x = sqrt(21)/2 - 1/2 or x = -1/2 - sqrt(21)/2

Step-by-step explanation:

Solve for x over the real numbers:

(3 x^3 + 4 x^2) - (2 x^3 + 3 x^2 - x) = (3 3 x + 4 2 x) - (2 3 x + 3 2 x - x)

(3 x^3 + 4 x^2) - (2 x^3 + 3 x^2 - x) = x^3 + x^2 + x and (3 x 3 + 4 x 2) - (2 x 3 + 3 x 2 - x) = 6 x:

x^3 + x^2 + x = 6 x

Subtract 6 x from both sides:

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

Factor x from the left hand side:

x (x^2 + x - 5) = 0

Split into two equations:

x = 0 or x^2 + x - 5 = 0

Add 5 to both sides:

x = 0 or x^2 + x = 5

Add 1/4 to both sides:

x = 0 or x^2 + x + 1/4 = 21/4

Write the left hand side as a square:

x = 0 or (x + 1/2)^2 = 21/4

Take the square root of both sides:

x = 0 or x + 1/2 = sqrt(21)/2 or x + 1/2 = -sqrt(21)/2

Subtract 1/2 from both sides:

x = 0 or x = sqrt(21)/2 - 1/2 or x + 1/2 = -sqrt(21)/2

Subtract 1/2 from both sides:

x = 0 or x = sqrt(21)/2 - 1/2 or x = -1/2 - sqrt(21)/2

assuming you mean addition: addition and subtraction are reverse operations.

step-by-step explanation:

step-by-step explanation:

x^2 -12x + 36 + y^2 = 36

x^2 + y^2 = 12x

r^2 = 12rcos(theta)

r = 12 cos(theta)