Answer:
[tex]\large\boxed{x=-2\ or\ x=\dfrac{1}{2}}[/tex]
Step-by-step explanation:
[tex]\text{The quadratic formula of an equation}\ ax^2+bx+c=0\\\\x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\\text{We have}\ 4x^2+6x-4=0\to a=4,\ b=6,\ c=-4.\\\\\text{Substitute:}\\\\x=\dfrac{-6\pm\sqrt{6^2-4(4)(-4)}}{2(4)}=\dfrac{-6\pm\sqrt{36+64}}{8}=\dfrac{-6\pm\sqrt{100}}{8}\\\\x=\dfrac{-6-10}{8}=\dfrac{-16}{8}=-2\ or\ x=\dfrac{-6+10}{8}=\dfrac{4}{8}=\dfrac{1}{2}[/tex]
