Find the roots of the following Quadratic Equations by applying quadratic formula.
(i) \(2x^2 - 7x + 3 = 0\)
(ii) \(2x^2 + x - 4 = 0\)
(iii) \(4x^2 + 4\sqrt{3}x + 3 = 0\)
(iv) \(2x^2 + x + 4 = 0\)

(i) \(2x^2 - 7x + 3 = 0\)

The general form of equation is \(ax^2 + bx + c = 0\)

Quadratic formula = \(x = {{-b ± \sqrt{b^2 - 4ac}} \over {2a}}\)

\(\Rightarrow \) \(x = {{7 ± \sqrt{(-7)^2 - 4(2)(3)}} \over {(2)(2)}}\)
\(\Rightarrow \)\(x = {{7 ± \sqrt{49 - 24}} \over {4}}\)
\(\Rightarrow \)\(x = {{7 ± 5} \over {4}}\)
\(\Rightarrow \)\(x = {{7 + 5} \over {4}} , {{7 - 5} \over {4}}\)
\(\Rightarrow \)\(x = 3 , {{1} \over {2}}\)


(ii) \(2x^2 + x - 4 = 0\)

The general form of equation is \(ax^2 + bx + c = 0\)

Quadratic formula = \(x = {{-b ± \sqrt{b^2 - 4ac}} \over {2a}}\)

\(\Rightarrow \)\(x = {{-1 ± \sqrt{(1)^2 - 4(2)(-4)}} \over {(2)(2)}}\)
\(\Rightarrow \)\(x = {{-1 ± \sqrt{33}} \over {4}}\)
\(\Rightarrow \)\(x = {{-1 + \sqrt{33}} \over {4}} , {{-1 - \sqrt{33}} \over {4}}\)


(iii) \(4x^2 + 4\sqrt{3}x + 3 = 0\)

The general form of equation is \(ax^2 + bx + c = 0\)

Quadratic formula = \(x = {{-b ± \sqrt{b^2 - 4ac}} \over {2a}}\)

\(\Rightarrow \)\(x = {{-4\sqrt{3} ± \sqrt{(4\sqrt{3})^2 - 4(4)(3)}} \over {(2)(4)}}\)
\(\Rightarrow \)\(x = {{-4\sqrt{3} ± \sqrt{0}} \over {8}}\)
\(\Rightarrow \)\(x = {{-\sqrt{3}} \over {2}} , {{-\sqrt{3}} \over {2}}\)


(iv) \(2x^2 + x + 4 = 0\)

The general form of equation is \(ax^2 + bx + c = 0\)

Quadratic formula = \(x = {{-b ± \sqrt{b^2 - 4ac}} \over {2a}}\)

\(\Rightarrow \)\(x = {{-1 ± \sqrt{(1)^2 - 4(2)(-4)}} \over {(2)(2)}}\)
\(\Rightarrow \)\(x = {{-1 ± \sqrt{-31}} \over {4}}\)

since square root of negative number does not exist.

Therefore, there is no solution.