Question

What are the zeroes of y=x^2+14x+40

Answer 1

$y=x^2+14x+40$ 

$y-x^2-14x-40=0$ 

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

$x= \dfrac{14+2 \sqrt{y+9} }{-2} , \dfrac{14-2 \sqrt{y+9} }{-2}$ 

$x=-7- \sqrt{y+9},-7+ \sqrt{y+9}$