1. **State the problem:** Solve for $x_{1/2}$ using the quadratic formula given by $$x_{1/2} = \frac{-0.7 \pm \sqrt{0.49 + 0.6}}{0.4}$$ 2. **Recall the quadratic formula:** For a quadratic equation $ax^2 + bx + c = 0$, the solutions are $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ Here, the expression inside the square root is called the discriminant $\Delta = b^2 - 4ac$. 3. **Calculate the discriminant:** $$0.49 + 0.6 = 1.09$$ So, $$x_{1/2} = \frac{-0.7 \pm \sqrt{1.09}}{0.4}$$ 4. **Evaluate the square root:** $$\sqrt{1.09} \approx 1.04403$$ 5. **Find the two solutions:** $$x_1 = \frac{-0.7 + 1.04403}{0.4}$$ $$x_2 = \frac{-0.7 - 1.04403}{0.4}$$ 6. **Simplify each fraction:** For $x_1$: $$x_1 = \frac{0.34403}{0.4}$$ Show cancellation: $$x_1 = \frac{\cancel{0.34403}}{\cancel{0.4}}$$ Calculate: $$x_1 = 0.860075$$ For $x_2$: $$x_2 = \frac{-1.74403}{0.4}$$ Show cancellation: $$x_2 = \frac{\cancel{-1.74403}}{\cancel{0.4}}$$ Calculate: $$x_2 = -4.360075$$ 7. **Final answer:** $$x_1 \approx 0.86, \quad x_2 \approx -4.36$$ These are the two solutions to the quadratic equation.