Solve the quadratic equation: 6(1 − x2) = 5x

The given equation is $6(1 - x^2) = 5x$.
Distribute the 6 on the left side:

$$6 - 6x^2 = 5x$$

Move all terms to one side to get $ax^2 + bx + c = 0$:

$$0 = 6x^2 + 5x - 6$$ $$6x^2 + 5x - 6 = 0$$

We need two numbers that multiply to $6 \times (-6) = -36$ and add up to 5. These numbers are 9 and $-4$.
Rewrite the middle term $5x$ as $9x - 4x$:

$$6x^2 + 9x - 4x - 6 = 0$$

Group the terms and factor by grouping:

$$(6x^2 + 9x) - (4x + 6) = 0$$

Factor out common terms from each group:

$$3x(2x + 3) - 2(2x + 3) = 0$$

Factor out the common binomial $(2x + 3)$:

$$(2x + 3)(3x - 2) = 0$$

Set each factor equal to zero and solve for $x$:

$$2x + 3 = 0 \quad \text{or} \quad 3x - 2 = 0$$ $$2x = -3 \quad \text{or} \quad 3x = 2$$ $$x = -\frac{3}{2} \quad \text{or} \quad x = \frac{2}{3}$$

The solutions for $x$ are $\boxed{x = -\frac{3}{2} \text{ or } x = \frac{2}{3}}$.