Question

p²+6p+5=0 help help :c

Answer 1

Vamos lá....

Equação do segundos grau:

P²+6p+5=0

Δ = $b^{2}$ - 4.a.c

Δ = $6^{2}$ - 4 . 1 . 5

Δ = 36 - 20 

Δ = 16

Agora a fórmula de Bhaskara:

P = (-b +/-  √Δ)/2a

P = $\frac{- 6 +/- \sqrt{16} }{2 . 1}$

P = $\frac{- 6 +/- 4}{2}$

P' = $\frac{-6 + 4 }{2} = \frac{-2}{2} = -1$

P'' = $\frac{-6 - 4}{2} = \frac{-10}{2} = -5$

Solução: {-5,-1}

Answer 2

Olá.

Vamos resolver como equação de segundo grau. 
$\mathsf{Usarem\ a\ f\'ormula:\ }\boxed{\mathsf{p=\dfrac{-b\pm\sqrt{b^2-4\cdot a\cdot c}}{2\cdot a}}}}\\\\\\\mathsf{a=1}\\\mathsf{b=6\ \ \ \ \ \ \ \ \ \ \boxed{\mathbf{p^2+6p+5=0}}}\\\mathsf{c=5}\\\\\mathsf{p=\dfrac{-6\pm\sqrt{6^2-4\cdot1\cdot5}}{2\cdot1}}\\\\\\\mathsf{p=\dfrac{-6\pm\sqrt{36-4\cdot5}}{2}}\\\\\\ \mathsf{p=\dfrac{-6\pm\sqrt{36-20}}{2}}\\\\\\ \mathsf{p=\dfrac{-6\pm\sqrt{16}}{2}}\\\\\\ \mathsf{p=\dfrac{-6\pm4}{2}}$

Vamos encontrar p' e p'':

$\begin{array}{ccc}\mathsf{p'=\dfrac{-6+4}{2}}&\ \ \ \ \ \ |\ \ \ \ \ \ &\mathsf{p''=\dfrac{-6-4}{2}}\\&\ \ \ \ \ \ |\ \ \ \ \ \ &\\\mathsf{p'=\dfrac{-2}{2}}&\ \ \ \ \ \ |\ \ \ \ \ \ &\mathsf{p''=\dfrac{-10}{2}}\\&\ \ \ \ \ \ |\ \ \ \ \ \ &\\\mathsf{p'=-1}&\ \ \ \ \ \ |\ \ \ \ \ \ &\mathsf{p''=-5}\\------------&\ \ \ \ \ \ |\ \ \ \ \ \ &------------\end{array}\\\\\\\\\\\mathsf{S=\{-5,\ -1\}}$

Qualquer dúvida, deixe nos comentários.
Bons estudos.