Solve for r\displaystyle {r}r within the complex number system.
(2r2+12r+16)(3r2−8r+4)−(7r2−28)=0\displaystyle {\left({2}{r}^{{2}}+{12}{r}+{16}\right)}{\left({3}{r}^{{2}}-{8}{r}+{4}\right)}-{\left({7}{r}^{{2}}-{28}\right)}={0}(2r2+12r+16)(3r2−8r+4)−(7r2−28)=0
r=\displaystyle {r}=r= Preview Question 6
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