Respuesta :

[tex] y=x^2+14x+40[/tex] 

[tex]y-x^2-14x-40=0[/tex] 

In general, given [tex]a{x}^{2}+bx+c=0[/tex], there exists two solutions where

[tex]x= \dfrac{14+2 \sqrt{y+9} }{-2} , \dfrac{14-2 \sqrt{y+9} }{-2}[/tex] 

[tex]x=-7- \sqrt{y+9},-7+ \sqrt{y+9} [/tex]