Which equation has the solutions x = -3 ± √3i/2 ?2x2 + 6x + 9 = 0x2 + 3x + 12 = 0x2 + 3x + 3 = 02x2 + 6x + 3 = 0

Answer:Answer is option C : [[tex]x^{2}[/tex] + 3x + 3 ] =0Note:  None of options matches with given question.instead of "-3" , there should be "-[tex]\frac{3}{2}[/tex]".Step-by-step explanation:Note:  None of options matches with given question.instead of "-3" , there should be "[tex]\frac{3}{2}[/tex]".  Here, First thing you have to observe the nature of roots.∴ x = -[tex]\frac{3}{2}[/tex]+[tex]\frac{\sqrt{3}}{2}[/tex]i and x = -[tex]\frac{3}{2}[/tex]-[tex]\frac{\sqrt{3}}{2}[/tex]∴ [ x+([tex]\frac{3}{2}[/tex]-[tex]\frac{\sqrt{3}}{2}[/tex]i) ][ x+([tex]\frac{3}{2}[/tex]+[tex]\frac{\sqrt{3}}{2}[/tex]i) ]=0∴ [ [tex]x^{2}[/tex] + x([tex]\frac{3}{2}[/tex]+[tex]\frac{\sqrt{3}}{2}[/tex]i)+ x([tex]\frac{3}{2}[/tex]-[tex]\frac{\sqrt{3}}{2}[/tex]i) + ([tex]\frac{3}{2}[/tex]-[tex]\frac{\sqrt{3}}{2}[/tex]i)([tex]\frac{3}{2}[/tex]+[tex]\frac{\sqrt{3}}{2}[/tex]i) ]=0∴ [[tex]x^{2}[/tex] + [tex]\frac{3}{2}[/tex]x + [tex]\frac{\sqrt{3}}{2}[/tex]ix + [tex]\frac{3}{2}[/tex]x - [tex]\frac{\sqrt{3}}{2}[/tex]ix + (3-[tex]\frac{\sqrt{3}}{2}[/tex]i)(3+[tex]\frac{\sqrt{3}}{2}[/tex]i) ] =0∴ [[tex]x^{2}[/tex] + 3x + ([tex]\frac{3}{2}[/tex]-[tex]\frac{\sqrt{3}}{2}[/tex]i)([tex]\frac{3}{2}[/tex]+[tex]\frac{\sqrt{3}}{2}[/tex]i) ] =0∴ [[tex]x^{2}[/tex] + 3x + [tex]\frac{9}{4}[/tex] - ([tex]\frac{\sqrt{3}}{2}[/tex]i)([tex]\frac{\sqrt{3}}{2}[/tex]i) ] =0∴ [[tex]x^{2}[/tex] + 3x + [tex]\frac{9}{4}[/tex] - ([tex]\frac{3}{4}[/tex]) [tex]i^{2}[/tex] ] =0∴ [[tex]x^{2}[/tex] + 3x + [tex]\frac{9}{4}[/tex] + ([tex]\frac{3}{4}[/tex]) ] =0∴ [[tex]x^{2}[/tex] + 3x + [tex]\frac{12}{4}[/tex] ] =0  ∴ [[tex]x^{2}[/tex] + 3x + 3 ] =0  Thus, Answer is option C : [[tex]x^{2}[/tex] + 3x + 3 ] =0