1. **State the problem:** Solve the quadratic equation $$2x^2 + 3x = 10$$ for $x$, with answers correct to 2 decimal places. 2. **Rewrite the equation:** Move all terms to one side to set the equation to zero: $$2x^2 + 3x - 10 = 0$$ 3. **Identify coefficients:** Here, $a = 2$, $b = 3$, and $c = -10$. 4. **Use the quadratic formula:** $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ 5. **Calculate the discriminant:** $$b^2 - 4ac = 3^2 - 4 \times 2 \times (-10) = 9 + 80 = 89$$ 6. **Calculate the roots:** $$x = \frac{-3 \pm \sqrt{89}}{4}$$ 7. **Evaluate the square root and simplify:** $$\sqrt{89} \approx 9.433$$ 8. **Calculate each root:** $$x_1 = \frac{-3 + 9.433}{4} = \frac{6.433}{4} = 1.61$$ $$x_2 = \frac{-3 - 9.433}{4} = \frac{-12.433}{4} = -3.11$$ 9. **Final answer:** $$x = 1.61 \text{ or } -3.11$$ The larger value is $1.61$ and the smaller is $-3.11$.