Question

6r^2+6r+12=0
solve the square​

Answer 1

`\large\boxed{\bold{NO\ REAL\ SOLUTION}}\\\text{in the set of complex numbers}\\\boxed{x=(-1\pm i\sqrt7)/(2(1))}`

Explanation:

`6r^2+6r+12=0\qquad\text{divide both sides by 6}\\\\(6r^2)/(6)+(6r)/(6)+(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=(-b)/(2a)\\\text{if}\ b^2-4ac>0,\ \text{hen the equation has two real solution}\ x=(-b\pm√(b^2-4ac))/(2a)`

`\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}`

`\text{If you want solution in the set of complex numbers, then}\\\\√(-7)=√((-1)(7))=√(-1)\cdot\sqrt7=i\sqrt7\\\\x=(-1\pm i\sqrt7)/(2(1))\\\\x=(-1\pm i\sqrt7)/(2)`