\[2x^2(4x^2  5) = 18\]
if x > 0, what is the solution to the equation above?
Solution:
Given equation:

\[2x^2(4x^2  5) = 18\]

Distribute:

\[8x^4 10x^2 = 18\]

Subtract 18:

\[8x^4 10x^2  18 = 0\]

Divide by 2:

\[4x^4 5x^2  9 = 0\]

Substitue y = x^2:

\[4y^2 5y  9 = 0\]

Factor:

\[(4y  9)(y + 1)\]

Solve for Zero Product Property: 4y  9 or y + 1 = 0; therefore, y = 9/4 or y = 1
Substitute x^2:

\[x^2 = 9/4; x^2 = 1\]

Since x is a real number, x^2 = 1 has no solutions. And if x must by positive, then the solution to x^2 = 9/4 is x = 3/2 or x = 1.5.
Final Answer: x = 1.5