Question

tem como fazer esse trinomio? y²+5y+10

Answer 1

$y^2 + 5y + 10 = 0$

Resolvendo por Bháskara:

a=1, b=5, c=−10

Δ=b2−4ac
Δ=(5)2−4*(1)*(−10)
Δ=25+40
Δ=65

$y = \dfrac{-b \pm \sqrt{\triangle}}{2*a} \\ \\ \\ y = \dfrac{-5 \pm \sqrt{65}}{2*1} \\ \\ \\ y = \dfrac{-5 \pm \sqrt{65}}{2} \\ \\ \\ y' = \dfrac{-5 + \sqrt{65}}{2} \\ \\ \\ y'' = \dfrac{-5 - \sqrt{65}}{2}$


S = {$\dfrac{-5 + \sqrt{65}}{2} ,\ \ \dfrac{-5 - \sqrt{65}}{2}$}