Answer:
[tex]\bold{k=\pm\dfrac{\sqrt6}{2}\approx\pm1.22}[/tex]
Step-by-step explanation:
[tex]2k^2-3=0\\\\a=2\,,\ \ b=0\,,\ \ c=-3\\\\k=\dfrac{-0\pm\sqrt{0^2-4\cdot2\cdot(-3)}}{2\cdot2}=\dfrac{\pm\sqrt{24}}{2\cdot2}=\dfrac{\pm2\sqrt{6}}{2\cdot2}=\pm\dfrac{\sqrt6}{2}\approx\pm1.22[/tex]
