1. **State the problem:** Solve the equation $$\frac{24}{x^2 + 2x - 8} - \frac{15}{x^2 + 2x - 3} = 2$$ for the variable $x$.\n\n2. **Factor the denominators:**\n\- Factor $x^2 + 2x - 8$: find factors of $-8$ that add to $2$, which are $4$ and $-2$, so $$x^2 + 2x - 8 = (x + 4)(x - 2).$$\n\- Factor $x^2 + 2x - 3$: find factors of $-3$ summing to $2$, which are $3$ and $-1$, so $$x^2 + 2x - 3 = (x + 3)(x - 1).$$\n\n3. **Rewrite the equation with factored denominators:**\n$$\frac{24}{(x + 4)(x - 2)} - \frac{15}{(x + 3)(x - 1)} = 2.$$\n\n4. **Find the common denominator:**\nThe denominators $(x + 4)(x - 2)$ and $(x + 3)(x - 1)$ share no factors, so the common denominator is the product:\n$$((x + 4)(x - 2))((x + 3)(x - 1)).$$\n\n5. **Rewrite each fraction with the common denominator:**\n\n- Multiply numerator and denominator of the first fraction by $(x + 3)(x - 1)$:\n$$\frac{24 (x + 3)(x - 1)}{(x + 4)(x - 2)(x + 3)(x - 1)}$$\n\n- Multiply numerator and denominator of the second fraction by $(x + 4)(x - 2)$:\n$$\frac{15 (x + 4)(x - 2)}{(x + 3)(x - 1)(x + 4)(x - 2)}.$$\n\n6. **Express original equation with common denominator:**\n$$\frac{24(x + 3)(x - 1) - 15(x + 4)(x - 2)}{(x + 4)(x - 2)(x + 3)(x - 1)} = 2.$$ 7. **Cross multiply to eliminate denominator:**\n$$24(x + 3)(x - 1) - 15(x + 4)(x - 2) = 2 (x + 4)(x - 2)(x + 3)(x - 1).$$\n\n8. **Expand each part:**\n\- Expand $24(x + 3)(x - 1)$:\n\quad $(x + 3)(x - 1) = x^2 + 2x - 3$\n\quad So, $24(x^2 + 2x - 3) = 24x^2 + 48x - 72$.\n\- Expand $15(x + 4)(x - 2)$:\n\quad $(x + 4)(x - 2) = x^2 + 2x - 8$\n\quad So, $15(x^2 + 2x - 8) = 15x^2 + 30x - 120$.\n\nLeft side after subtraction:\n$$24x^2 + 48x - 72 - (15x^2 + 30x - 120) = (24x^2 - 15x^2) + (48x - 30x) + (-72 + 120) = 9x^2 + 18x + 48.$$ 9. **Expand the right side:**\nExpand $(x + 4)(x - 2)(x + 3)(x - 1)$. Group in pairs:\n\- $(x + 4)(x - 2) = x^2 + 2x - 8$\n\- $(x + 3)(x - 1) = x^2 + 2x - 3$\nMultiply these:\n$$ (x^2 + 2x - 8)(x^2 + 2x - 3). $$\nMultiply term by term:\n$$ x^2(x^2 + 2x - 3) + 2x(x^2 + 2x - 3) - 8(x^2 + 2x - 3) = $$\n$$ x^4 + 2x^3 - 3x^2 + 2x^3 + 4x^2 - 6x - 8x^2 - 16x + 24 = $$\nSimplify:\n$$ x^4 + (2x^3 + 2x^3) + (-3x^2 + 4x^2 - 8x^2) + (-6x - 16x) + 24 = $$\n$$ x^4 + 4x^3 - 7x^2 - 22x + 24.$$\n\nMultiply by 2:\n$$2x^4 + 8x^3 - 14x^2 - 44x + 48.$$\n\n10. **Set equation:**\n$$9x^2 + 18x + 48 = 2x^4 + 8x^3 - 14x^2 - 44x + 48.$$\n\n11. **Bring all terms to one side:**\n$$0 = 2x^4 + 8x^3 - 14x^2 - 44x + 48 - 9x^2 -18x - 48,$$\nwhich simplifies to\n$$0 = 2x^4 + 8x^3 - 23x^2 - 62x.$$\n\n12. **Factor the polynomial:** First factor out $x$:\n$$x(2x^3 + 8x^2 - 23x - 62) = 0.$$\n\n13. **Check the first solution:**\n$x = 0$ is one root. Check if it causes division by zero in original denominators:\n\- Denominator 1 at $x=0$: $0^2 + 2\cdot0 - 8 = -8 \neq 0$\n\- Denominator 2 at $x=0$: $0^2 + 2\cdot0 - 3 = -3 \neq 0$\nSo $x=0$ is valid.\n\n14. **Solve cubic equation:**\n$$2x^3 + 8x^2 - 23x - 62 = 0.$$\n\nTry rational roots using the Rational Root Theorem with possible roots from factors of 62 over 2: $\pm1, \pm2, \pm31, \pm62, \pm\frac{1}{2}, \pm\frac{31}{2}$.\n\nTest $x = 2$:\n$2(2)^3 + 8(2)^2 - 23(2) - 62 = 2(8) + 8(4) - 46 - 62 = 16 + 32 - 46 - 62 = 48 - 108 = -60 \neq 0.$\n\nTest $x = -2$:\n$2(-2)^3 + 8(-2)^2 - 23(-2) - 62 = 2(-8) + 8(4) + 46 - 62 = -16 +32 +46 -62 = 0.$\nSo $x = -2$ is a root.\n\n15. **Factor out $(x + 2)$:**\nDivide $2x^3 + 8x^2 - 23x - 62$ by $(x + 2)$ using synthetic or polynomial division:\nQuotient is $2x^2 + 4x -31$.\n\n16. **Solve quadratic:**\n$$2x^2 + 4x - 31 = 0.$$\nUse quadratic formula:\n$$x = \frac{-4 \pm \sqrt{4^2 - 4(2)(-31)}}{2 \cdot 2} = \frac{-4 \pm \sqrt{16 + 248}}{4} = \frac{-4 \pm \sqrt{264}}{4}.$$\nSimplify $\sqrt{264} = \sqrt{4 \cdot 66} = 2\sqrt{66}$, so\n$$x = \frac{-4 \pm 2\sqrt{66}}{4} = \frac{-2 \pm \sqrt{66}}{2}.$$\n\n17. **Collect all solutions:**\n$$x = 0, x = -2, x = \frac{-2 + \sqrt{66}}{2}, x = \frac{-2 - \sqrt{66}}{2}.$$\n\n18. **Check for excluded values:**\nOriginal denominators zero if\n$$x^2 + 2x - 8 = 0 \implies (x+4)(x-2)=0 \implies x = -4, 2,$$\n$$x^2 + 2x - 3 = 0 \implies (x+3)(x-1) = 0 \implies x = -3, 1.$$ None of these values are solutions found, so all are valid.\n\n**Final answer:**\n$$x = 0, \quad x = -2, \quad x = \frac{-2 + \sqrt{66}}{2}, \quad x = \frac{-2 - \sqrt{66}}{2}.$$