Question

Mim ajudem por favor​

Answer 1

${x}^{2} + 2x + 3 = 0 \\ (x + 2)(x + 1) = 0 \\ x = - 2 \\ ou\\ x = - 1$

B)

$2 {x}^{2} + 4x + 6 = 0 \\ Δ = b { }^{2} - 4ac \\ Δ = {4}^{2} - 4 \times 2 \times 6 \\ Δ = 16 - 48 \\ Δ = - 32$

Não tem Raízes

C)

$5x {}^{2} - 2x + 3 \\ Δ = b {}^{2} - 4ac \\Δ = ( - 2) {}^{2} - 4 \times 5 \times 3 \\ Δ = 4 + 60 \\ Δ = 64 \\ \\ x = \frac{ - ( - 2) + \sqrt{64} }{10} = \frac{2 + 8}{10} = \frac{10}{10} = 1 \\ \\ x = \frac{2 - 8}{10} = \frac{6}{10} = \frac{3}{5}$

D)

${x}^{2} - x + 1 = 0 \\ Δ = ( - 1) {}^{2} - 4 \times 1 \times 1 \\ Δ = 1 - 4 = - 3$

Não existe

E)

$2x {}^{2} + x - 3 = 0 \\ Δ = 1 {}^{2} - 4 \times 2 \times ( - 3) \\ Δ = 1 + 24 = 25 \\ \\ x = \frac{ 1 + \sqrt{25} }{2 \times 2} = \frac{1 + 5}{4} = \frac{6}{4} = \frac{3}{2} \\ \\ x = \frac{1 - 5}{4} = - \frac{4}{4} = - 1$