Question

quero saber como resolver essa matemática

x² + x - 12 =0

-x² + 14 x - 49 =0

x² - 2 x + 5 =0

Answer 1

x² + x -12 =0
Δ = 1² - 4(1)(-12)
Δ = 1 + 48
Δ = 49
$x = \frac{-1\pm7}{2} \\ x' = 3 \\ x'' = -4$

x² + 14x - 49 = 0
Δ = (14)² - 4(1)(-49)
Δ = 196 + 196
Δ = 392
$x = \frac{14\pm14 \sqrt{2} }{2} \\ x = \frac{ 2(7 \pm 7 \sqrt{2} }{2} \\ x = 7\pm7 \sqrt{2}$

x² - 2x + 5 = 0
Δ = (-2)² - 4(1)(5)
Δ = 4 - 20
Δ = -16
S = {x ∉ |R}