(i) $2x^{2}$ – 7x + 3 = 0
On comparing the given equation with $ax^{2} + bx + c $= 0, we get,
a = 2, b = -7 and c = 3
By using the quadratic formula, we get,
x=$\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
⇒ x = $\dfrac{(7\pm\sqrt{(49 – 24)})}{4}$
⇒ x = $\dfrac{(7±\sqrt{25})}{4}$
⇒ x = $\dfrac{(7±5)}{4}$
⇒ x = $\dfrac{(7+5)}{4}$or x = $\dfrac{(7-5)}{4}$
⇒ x = $ \dfrac{12}{4}$ or $ \dfrac{2}{4} $
∴ x = 3 or $\dfrac{1}{2}$
(ii) 2x2 + x – 4 = 0
On comparing the given equation with $ax^{2} + bx + c$ = 0, we get,
a = 2, b = 1 and c = -4
By using the quadratic formula, we get,
x=$\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
⇒x = $\dfrac{(-1±\sqrt{1+32})}{4}$
⇒x = $\dfrac{(-1±\sqrt{33})}{4}$
∴ x = $\dfrac{(-1+\sqrt{33})}{4}$ or x = $\dfrac{(-1-\sqrt{33})}{4}$
(iii) 4x2 + $4\sqrt{3}x$ + 3 = 0
On comparing the given equation with $ax^{2} + bx + c$ = 0, we get,
a = 4, b = $4\sqrt{3}$ and c = 3
By using the quadratic formula, we get,
x=$\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
⇒ x = $\dfrac{(-4\sqrt{3}±\sqrt{48-48})}{8}$
⇒ x = $\dfrac{(-4\sqrt{3}±0)}{8}$
∴ x = $\dfrac{-\sqrt{3}}{2} or x = \dfrac{-\sqrt{3}}{2}$
(iv) 2x2 + x + 4 = 0
On comparing the given equation with $ax^{2} + bx + c$ = 0, we get,
a = 2, b = 1 and c = 4
By using the quadratic formula, we get
x=$\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
⇒ x = $\dfrac{(-1±\sqrt{1-32})}{4}$
⇒ x = $\dfrac{(-1±\sqrt{-31})}{4}$
As we know, the square of a number can never be negative. Therefore, there is no real solution for the given equation.