15-04-2017
Mathematics

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

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Аноним Аноним

15-04-2017

[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]

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