Question

resolva as equações de 2 grau em IR ultiliando a formula resolutiva OBS:APENAS NÚMERO 1

Answer 1

Eu vou fazer direto

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${x}^{2} - 7x + 12 = 0 \\ Δ = {( - 7)}^{2} - 4 \times 1 \times 12 \\ Δ = 49 - 48 \\ Δ =1 \\ \\ x \frac{7±1}{2} \: x^{1} = \frac{8}{2} = 4 \: \: \: \: \: {x}^{2} = \frac{6}{2} = 3$

${x}^{2} + 4x - 12 = 0 \\ Δ = {4}^{2} - 4 \times 1 \times ( - 12) \\ Δ = 16 + 48 \\ Δ = 64 \\ \\ x \frac{ - 4±8}{2} \: \: \: \: {x}^{1} = \frac{ - 4 + 8}{2} = \frac{4}{2} = 2 \: \: \: \: \\ {x}^{2} = \frac{ - 4 - 8}{2} = \frac{ - 12}{2} = - 6$

${x}^{2} + 6x + 5 = 0 \\ Δ = {6}^{2} - 4 \times 1 \times 5 \\ Δ = 36 - 20 \\ Δ = 16 \\ \\ x \frac{ - 6±4}{2} \\ {x}^{1} = \frac{ - 6 + 4}{2} = \frac{ - 2}{2} = - 2 \: \: \: \\ {x}^{2} = \frac{ - 6 - 4}{2} = \frac{ - 10}{2} = - 5$

${x}^{2} - 12x + 36 \\ Δ = {( - 12)}^{2} - 4 \times 1 \times 36 \\ Δ = 144 - 144\\ Δ = 0 \\ \\ x \frac{12±0}{2} \: \: \: \: \: x = \frac{12}{2} = 6$

${x}^{2} + 2x + 1 = 0 \\ Δ = {2}^{2} - 4 \times 1 \times 1 \\ Δ =4 - 4 \\ Δ =0 \\ \\ x \frac{ - 2±0}{2} \: \: \: \: \: x = \frac{ - 2}{2} = - 1$

$2x^2 + x + 3 \\ Δ =1^{2} - 4 \times 2 \times 3 \\ Δ =1 - 24 \\ Δ = - 23 \\ \\ x \frac{ - 1±23}{4} \\ {x}^{1} = \frac{ - 1 + 23}{4} = \frac{22}{4} = \frac{11}{2} \\ {x}^{2} = \frac{ - 1 - 23}{4} = \frac{ - 24}{4} = - 5$

Estão aí as 6 letras, espero ter ajudado