Where’s the triangle, isn’t there supposed to be a picture with the question
Compose the quadratic condition in standard shape, ax2 + bx + c = 0. Recognize the values of a, b, c. Write the Quadratic Equation. At that point substitute within the values of a, b, c. Simplify. Check the arrangements.
do you need to include the wiggle infront of the first bracket? if not;
(x+1) ÷ [(x^2+2) x (2x-3dx)]
x^2 x 2x = 2x^3
x^2 x -3dx = -3dx^3
2 x 2x = 4x
2 x -3dx = - 6dx
i cant find a way to make it equal 0 so i think the answer is just
x+1 over 2x^3 - 3dx^3 + 4x - 6dx as a fraction
Let the number be x.
x*x*x = 54
x³ = 54 Taking cube roots
x = ∛54
x ≈ 3.780
