Question

Preciso resolver 25y^2+10y-1=0 preciso pra hj

Answer 1

$25y^2+10y-1=0 \\ \\ a=25 \\ b=10 \\ c=-1 \\ \\ x= \frac{ -b \frac{+}{} \sqrt{b^{2}-4.a.c}}{2a} \\ \\ x= \frac{-10 \frac{+}{} \sqrt{625+100} }{2.25} \\ \\ x= \frac{-10 \frac{+}{} \sqrt{725} }{50} \\ \\ x'= \frac{-10+ 5\sqrt{29} }{50} = \frac{-10}{50} + \frac{ 5\sqrt{29} }{50} =- \frac{1}{5} + \frac{ \sqrt{29}}{10} = \frac{-2+ \sqrt{29 } }{10} \\ \\ x''= \frac{-10- 5\sqrt{29} }{50} = \frac{-2- \sqrt{29} }{10} \\ \\ S=( \frac{-2- \sqrt{29}}{10} ; \frac{-2+ \sqrt{29}}{10})$