Answer:
[tex]\large\boxed{\bold{NO\ REAL\ SOLUTION}}\\\text{in the set of complex numbers}\\\boxed{x=\dfrac{-1\pm i\sqrt7}{2(1)}}[/tex]
Step-by-step explanation:
[tex]6r^2+6r+12=0\qquad\text{divide both sides by 6}\\\\\dfrac{6r^2}{6}+\dfrac{6r}{6}+\dfrac{12}{6}=0\\\\r^2+r+2=0\\\\\text{Use the quadratic formula:}\\\\\text{for}\ ax^2+bx+c=0\\\\\text{if}\ b^2-4ac<0,\ \text{hen the equation has no real solution}\\\text{if}\ b^2-4ac=0,\ \text{hen the equation has one real solution}\ x=\dfrac{-b}{2a}\\\text{if}\ b^2-4ac>0,\ \text{hen the equation has two real solution}\ x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
[tex]\text{We have}\ a=1,\ b=1,\ c=2.\\\\b^2-4ac=1^2-(4)(1)(2)=1-8=-7<0\\\\\bold{NO\ REAL\ SOLUTION}[/tex]
[tex]\text{If you want solution in the set of complex numbers, then}\\\\\sqrt{-7}=\sqrt{(-1)(7)}=\sqrt{-1}\cdot\sqrt7=i\sqrt7\\\\x=\dfrac{-1\pm i\sqrt7}{2(1)}\\\\x=\dfrac{-1\pm i\sqrt7}{2}[/tex]
