Respuesta :

we can't facotr this
for
ax^2+bx+c=0
x=[tex] \frac{-b+/- \sqrt{b^2-4ac} }{2a} [/tex]
1x^2+6x+7
x=[tex] \frac{-6+/- \sqrt{6^2-4(1)(7)} }{2(1)} [/tex]
x=[tex] \frac{-6+/- \sqrt{36-28} }{2} [/tex]
x=[tex] \frac{-6+/- \sqrt{8} }{2} [/tex]
x=[tex] \frac{-6+/- 4\sqrt{2} }{2} [/tex]
x=[tex] -3+/- 2\sqrt{2} [/tex]


x=[tex] -3+ 2\sqrt{2} [/tex] and x=[tex]-3- 2\sqrt{2} [/tex]


Using the quadratic formula.
Hope this helps.
Ver imagen anthonyf2312