a: \(4x^2-9=0\)
=>(2x-3)(2x+3)=0
=>x=3/2 hoặc x=-3/2
b: \(5x^2+20=0\)
nên \(x^2+4=0\)(vô lý)
c: \(2x^2-2+\sqrt{3}=0\)
\(\Leftrightarrow2x^2=2-\sqrt{3}\)
\(\Leftrightarrow x^2=\dfrac{4-2\sqrt{3}}{4}\)
hay \(x\in\left\{\dfrac{\sqrt{3}-1}{2};\dfrac{-\sqrt{3}+1}{2}\right\}\)