Answer: [tex]a=\dfrac{1\pm \sqrt{13}}{6}[/tex]
Step-by-step explanation:
Given
Quadratic Equation is [tex]x^2-x-(2a+2)[/tex]
[tex]-4a[/tex] is the solution of the given equation
[tex]\therefore (-4a)^2-(-4a)-(2a+2)^2=0\\\Rightarrow 16a^2+4a-(4a^2+4+8a)=0\\\Rightarrow 12a^2-4a-4=0\\\Rightarrow 3a^2-a-1=0\\\\\Rightarrow a=\dfrac{1\pm \sqrt{1+12}}{2\times 3}=\dfrac{1\pm \sqrt{13}}{6}[/tex]
