Question

t² + 6 t+9 = 0 equação de 2grau

Answer 1

$t^{2} +6t + 9 = 0$

a=1
b=6
c=9

$\frac{-b \pm \sqrt{-b^2 -4*a*c}}{2*a}$

Vamos calcular o valor de delta
Δ=b2−4ac
Δ=(6)2−4*(1)*(9)
Δ=36−36
Δ=0

$x = \frac{-b \pm \sqrt{\triangle}}{2*a} \\ \\ x = \frac{-6 \pm \sqrt{0}}{2*1} \\ \\ x = \frac{-6 \pm {0}}{2} \\ \\ x' = \frac{-6 + {0}}{2} \\ \\ x' = \frac{-6}{2} \\ \\ x' = -3 \\ \\ \\ x '' = \frac{-6- {0}}{2} \\ \\ x' = \frac{-6}{2} \\ \\ x' = -3 \\ \\ \\ S = \left \{ - 3 \right. \}$