ak123
ak123
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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