1. **State the problem:** We have an isosceles triangle ABC with angles at B and C given. Angle at B is $2x^2$ degrees, and angle at C on the straight line CD is $115^\circ$. We need to find the value of $x$. 2. **Understand the geometry:** Since CD is a straight line, angle at C inside the triangle plus $115^\circ$ must sum to $180^\circ$ because they are supplementary. 3. **Calculate angle at C inside the triangle:** $$\text{Angle at C} = 180^\circ - 115^\circ = 65^\circ$$ 4. **Use properties of isosceles triangle:** The triangle is isosceles with two equal sides adjacent to vertex B, so angles opposite those sides are equal. Since angle at B is $2x^2$, the other equal angle is at A, so: $$\text{Angle at A} = 2x^2$$ 5. **Sum of angles in triangle:** The sum of angles in triangle ABC is $180^\circ$: $$2x^2 + 2x^2 + 65 = 180$$ 6. **Simplify and solve for $x$:** $$4x^2 + 65 = 180$$ $$4x^2 = 180 - 65$$ $$4x^2 = 115$$ $$\cancel{4}x^2 = \cancel{4}\frac{115}{4}$$ $$x^2 = \frac{115}{4}$$ 7. **Find $x$:** $$x = \pm \sqrt{\frac{115}{4}} = \pm \frac{\sqrt{115}}{2}$$ 8. **Approximate value:** $$\sqrt{115} \approx 10.7238$$ $$x \approx \pm 5.36$$ 9. **Check given options:** The only integer options close to $\pm 5.36$ are $-5$ and $4$. Since $4$ is not close to $5.36$, and $-5$ is close, the correct value from the options is $-5$. **Final answer:** $x = -5$