x^2+4x-y = 6
Subtract 6 from each side.
x^2+4x-y-6 = 0
x^2+4x+4-10-y = 0
Add 10,y to each side.
x^2+4x+4 = y+10
(x+2)^2 = y+10 --------> (1)
(x+2)^2+7y^2 = 5 -----------> (2)
From equation (1) substitute the value of (x+2)^2 in equation (2)
y+10+7y^2 = 5
Subtract 5 from each side.
7y^2+y +10-5 = 0
7y^2+y+5 = 0
Compare the quadratic equatiion to ax^2+bx+c = 0
Roots are x = (-b±√(b^2-4ac))/2a.
y = [-1±√1-(4*7*5)]/2*7
y = (-1±√1-140)/14
y = (-1±√-139)/14
negitive square root is not possible.
So the system has no solution.