Explicação passo-a-passo:
a)
[tex] {x}^{2} - x - 6 = 0 \\ x = \frac{1 + - \sqrt{ {( - 1)}^{2} - 4 \times 1 \times ( - 6) } }{2 \times 1} \\ \\ x = \frac{1 + - \sqrt{1 + 24} }{2} \\ \\ x = \frac{1 + - \sqrt{25} }{2 } \\ \\ x1 = \frac{1 + 5}{2} = \frac{6}{2} = 3 \\ \\ x2 = \frac{1 - 5}{2} = - \frac{ 4}{2} = - 2[/tex]
b)
[tex] - {x}^{2} - 4x + 5 = 0 \\ x = \frac{4 + - \sqrt{ {( - 4)}^{2} - 4 \times ( - 1) \times 5 } }{2 \times ( - 1)} \\ \\ x = \frac{4 + - \sqrt{16 + 20} }{ - 2} \\ \\ x = \frac{4 + - \sqrt{36} }{ - 2} \\ \\ x1 = \frac{4 + 6}{ - 2} = - \frac{10}{2} = - 5 \\ \\ x2 = \frac{4 - 6}{ - 2} = \frac{ - 2}{ - 2} = 1[/tex]
c)
[tex] {x}^{2} - 7x + 10 = 0 \\ x = \frac{7 + - \sqrt{ {( - 7)}^{2} - 4 \times 1 \times 10} }{2 \times 1} \\ \\ x = \frac{7 + - \sqrt{49 - 40} }{2} \\ \\ x = \frac{7 + - \sqrt{9} }{2} \\ \\ x1 = \frac{7 + 3}{2} = \frac{10}{2} = 5 \\ \\ x2 = \frac{7 - 3}{2} = \frac{4}{2} = 2[/tex]
d)
[tex]2 {x}^{2} + 2x - 12 = 0 \\ {x}^{2} + x - 6 = 0 \\ x = \frac{ - 1 + - \sqrt{ {1}^{2} - 4 \times 1 \times ( - 6) } }{2 \times 1} \\ \\ x = \frac{ - 1 + - \sqrt{1 + 24} }{2} \\ \\ x = \frac{ - 1 + - \sqrt{25} }{2} \\ \\ x1 = \frac{ - 1 + 5}{2} = \frac{4}{2} = 2 \\ \\ x2 = \frac{ - 1 - 5}{2} = - \frac{6}{2} = - 3[/tex]
