Question

"Solving Equations by Completing the Square: m^2+2m-48=-6

Answer 1

$m^2+2m-48=-6 \\ \\ m^2+2m-48+6=0\\ \\m^2+2m-42=0\\ \\a=1, \ b= 2, \ c= -42 \\ \\\Delta = b^{2}-4ac = 2^{2}-4*1*(-42)= 4+168 =172 \\\\\sqrt{\Delta }=\sqrt{172} =\sqrt{4*43}=2\sqrt{43}\\ \\x_{1}=\frac{-b-\sqrt{\Delta }}{2a} =\frac{-2-2\sqrt{43}}{2}=\frac{2(-1-\sqrt{43})}{2}= -1-\sqrt{43} \\ \\x_{2}=\frac{-b+\sqrt{\Delta }}{2a} = \frac{-2+2\sqrt{43}}{2}=\frac{ 2(-1+\sqrt{43})}{2}= -1+\sqrt{43}$

Answer 2

$m^2+2m-48=-6\\\\m^2+2m\cdot1+1^2-1^2=-6+48\\\\(m+1)^2-1=42\\\\(m+1)^2=42+1\\\\(m+1)^2=43\iff m+1=-\sqrt{43}\ \vee\ m+1=\sqrt{43}\\\\m=-1-\sqrt{43}\ \vee\ m=-1+\sqrt{43}$