The general form of a quadratic equation is $$ax^2 + bx + c = 0$$, where a, b, and c are constants and $$a \ne 0$$.

Brahmagupta's Formula for solving quadratic equations is $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$, where a, b, and c are the coefficients of the quadratic equation.

The discriminant of a quadratic equation is $$b^2 - 4ac$$. It determines the nature of the roots of the equation.

The discriminant can be positive, negative, or zero. The value of the discriminant determines the nature of the roots of the equation.

If the discriminant is positive, the roots of the quadratic equation are real and distinct.

If the discriminant is negative, the roots of the quadratic equation are imaginary.

If the discriminant is zero, the roots of the quadratic equation are real and equal.

Using Brahmagupta's Formula, we have $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$. Substituting the values of a, b, and c, we get $$x = \frac{-4 \pm \sqrt{4^2 - 4(1)(3)}}{2(1)}$$. Simplifying this, we get $$x = -1 \pm \sqrt{2}$$. Therefore, the solution set is ({-1 \pm \sqrt{2})).

Using Brahmagupta's Formula, we have $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$. Substituting the values of a, b, and c, we get $$x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(2)(2)}}{2(2)}$$. Simplifying this, we get $$x = \frac{5 \pm \sqrt{21}}{4}$$. Therefore, the solution set is ({\frac{5 \pm \sqrt{21}}{4})).

Using Brahmagupta's Formula, we have $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$. Substituting the values of a, b, and c, we get $$x = \frac{-7 \pm \sqrt{7^2 - 4(3)(-6)}}{2(3)}$$. Simplifying this, we get $$x = \frac{-7 \pm \sqrt{85}}{6}$$. Therefore, the solution set is ({\frac{-7 \pm \sqrt{85}}{6})).

Using Brahmagupta's Formula, we have $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$. Substituting the values of a, b, and c, we get $$x = \frac{-(-12) \pm \sqrt{(-12)^2 - 4(4)(9)}}{2(4)}$$. Simplifying this, we get $$x = \frac{6 \pm \sqrt{15}}{2}$$. Therefore, the solution set is ({\frac{6 \pm \sqrt{15}}{2})).

Using Brahmagupta's Formula, we have $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$. Substituting the values of a, b, and c, we get $$x = \frac{-2 \pm \sqrt{2^2 - 4(5)(-3)}}{2(5)}$$. Simplifying this, we get $$x = \frac{-1 \pm \sqrt{23}}{5}$$. Therefore, the solution set is ({\frac{-1 \pm \sqrt{23}}{5})).

Using Brahmagupta's Formula, we have $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$. Substituting the values of a, b, and c, we get $$x = \frac{-(-11) \pm \sqrt{(-11)^2 - 4(6)(3)}}{2(6)}$$. Simplifying this, we get $$x = \frac{11 \pm \sqrt{109}}{12}$$. Therefore, the solution set is ({\frac{11 \pm \sqrt{109}}{12})).

Using Brahmagupta's Formula, we have $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$. Substituting the values of a, b, and c, we get $$x = \frac{-(-13) \pm \sqrt{(-13)^2 - 4(7)(6)}}{2(7)}$$. Simplifying this, we get $$x = \frac{13 \pm \sqrt{121}}{14}$$. Therefore, the solution set is ({\frac{13 \pm \sqrt{121}}{14})).