1. The problem is to solve the equation $$x^2(2x + 3) = 17x - 12$$. 2. First, expand the left side: $$x^2(2x) + x^2(3) = 2x^3 + 3x^2$$. 3. Rewrite the equation as: $$2x^3 + 3x^2 = 17x - 12$$. 4. Bring all terms to one side: $$2x^3 + 3x^2 - 17x + 12 = 0$$. 5. Next, try to factor the cubic polynomial by searching for rational roots using the Rational Root Theorem. Possible roots are factors of 12 over factors of 2: $$\pm1, \pm2, \pm3, \pm4, \pm6, \pm12, \pm\frac{1}{2}, \pm\frac{3}{2}, \pm\frac{...}{}$$. 6. Test $$x = 1$$: $$2(1)^3 + 3(1)^2 - 17(1) + 12 = 2 + 3 - 17 + 12 = 0$$, so $$x=1$$ is a root. 7. Divide the polynomial by $$x-1$$ using polynomial division or synthetic division: $$2x^3 + 3x^2 - 17x + 12 \div (x - 1) = 2x^2 + 5x - 12$$. 8. Solve the quadratic equation $$2x^2 + 5x - 12 = 0$$ using the quadratic formula: $$x = \frac{-5 \pm \sqrt{5^2 - 4(2)(-12)}}{2 \times 2} = \frac{-5 \pm \sqrt{25 + 96}}{4} = \frac{-5 \pm \sqrt{121}}{4}$$. 9. Simplify the square root: $$\sqrt{121} = 11$$. 10. Get the two roots from the quadratic: $$x = \frac{-5 + 11}{4} = \frac{6}{4} = 1.5$$, $$x = \frac{-5 - 11}{4} = \frac{-16}{4} = -4$$. 11. The solutions to the equation are $$x = 1$$, $$x = 1.5$$, and $$x = -4$$.