UBSTITUTION olve the equation given to find all real roots. \[ -z^{4}+6 z^{2}-35=-3 z^{4}-3 z^{2} \]

To solve the equation \(-z^{4}+6 z^{2}-35=-3 z^{4}-3 z^{2}\), first, let's simplify it. Bring all terms to one side: \[ -z^{4} + 6z^{2} - 35 + 3z^{4} + 3z^{2} = 0 \] This simplifies to: \[ 2z^{4} + 9z^{2} - 35 = 0 \] Next, let’s make a substitution to make it easier to solve. Let \(u = z^{2}\). The equation becomes: \[ 2u^{2} + 9u - 35 = 0 \] Now we can use the quadratic formula \(u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\): Here, \(a = 2\), \(b = 9\), and \(c = -35\). Calculating the discriminant: \[ D = b^{2} - 4ac = 9^{2} - 4 \cdot 2 \cdot (-35) = 81 + 280 = 361 \] Since \(D\) is positive, there are two real roots. Now calculating \(u\): \[ u = \frac{-9 \pm \sqrt{361}}{2 \cdot 2} = \frac{-9 \pm 19}{4} \] This gives us: 1. \(u = \frac{10}{4} = 2.5\) 2. \(u = \frac{-28}{4} = -7\) (not valid since \(u = z^2\) must be non-negative) Now substituting back \(u = z^2\): \[ z^{2} = 2.5 \implies z = \pm\sqrt{2.5} \] Thus, the real roots for \(z\) are: \[ z = \pm\frac{\sqrt{10}}{2} \] These are the solutions for the original equation.