To solve the equation \(2x^4 + 3x^2 - 5 = 0\), we can perform a substitution to simplify it. Let \(y = x^2\). This means \(x^4 = y^2\). The equation then becomes:

\[2y^2 + 3y - 5 = 0.\]

Now we need to solve this quadratic equation for \(y\). We apply the quadratic formula, where \(a = 2\), \(b = 3\), and \(c = -5\):

\[y = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}.\]

Calculate the discriminant:

\[b^2 - 4ac = 3^2 - 4 \times 2 \times (-5) = 9 + 40 = 49.\]

Since the discriminant is a perfect square, we can solve for \(y\):

\[y = \frac{{-3 \pm \sqrt{49}}}{4} = \frac{{-3 \pm 7}}{4}.\]

This gives us two solutions for \(y\):

1. \(y = \frac{{-3 + 7}}{4} = \frac{4}{4} = 1.\)
2. \(y = \frac{{-3 - 7}}{4} = \frac{-10}{4} = -\frac{5}{2}.\)

Since \(y = x^2\), and \(x^2\) cannot be negative, we discard \(y = -\frac{5}{2}\).

Using \(y = 1\), we find \(x^2 = 1\), which gives:

\[x = \pm 1.\]

Thus, the solutions for the original equation are \(x = 1\) and \(x = -1\).