Correct option is (B) Real and equal
The given quadratic equation is
\(\frac{4}{3} x^{2}-2 x+\frac{3}{4}=0\)
Here,
\(a=\frac{4}{3} , b=-2 \text { and } c=\frac{3}{4}\)
\(D =b^{2}-4 a c\)
\(=(-2)^{2}-4 \times \frac{4}{3} \times \frac{3}{4}\)
\(=4-4\)
\(=0\)
\(P =0\)
Hence, the equation has real and equal roots.