Solve the following equation using the method of 'completing the square': $4 x^{2}+3 x+5=0$

(D) Given equation: $4 x^{2}+3 x+5=0$
Divide the entire equation by $4$ to make the coefficient of $x^2$ equal to $1$:
$x^{2} + \frac{3}{4}x + \frac{5}{4} = 0$
To complete the square,add and subtract the square of half the coefficient of $x$,which is $(\frac{1}{2} \times \frac{3}{4})^2 = (\frac{3}{8})^2 = \frac{9}{64}$:
$x^{2} + \frac{3}{4}x + \frac{9}{64} - \frac{9}{64} + \frac{5}{4} = 0$
$(x + \frac{3}{8})^2 + \frac{-9 + 80}{64} = 0$
$(x + \frac{3}{8})^2 + \frac{71}{64} = 0$
$(x + \frac{3}{8})^2 = -\frac{71}{64}$
Since the square of any real number cannot be negative,there is no real value of $x$ that satisfies this equation. Thus,the equation has no real roots.