Question

-x(ao quadrado)+x+12=0 na fórmula de Bhaskara

Answer 1

$-x^2+x+12=0 \\ \\ \\ x_{1\ y\ 2} = \dfrac{-b\pm \sqrt{b^2-4ac} }{2a} \qquad a= -1\qquad b=1\qquad c= 12 \\ \\ \\ x_{1\ y\ 2} = \dfrac{-1\pm \sqrt{1-4(-1)(12)} }{2(-1)}\\ \\ \\ x_{1\ y\ 2} = \dfrac{-1\pm \sqrt{1+48} }{-2} \\ \\ \\ x_{1\ y\ 2} = \dfrac{-1\pm \sqrt{49} }{-2} \ \\ \\ \\ x_{1\ y\ 2} = \dfrac{-1\pm 7 }{-2} \\ \\\\ x_{1} = \dfrac{-1+7 }{-2} \qquad \qquad x_{2} = \dfrac{-1-7 }{-2} \\ \\ x_{1} = \dfrac{6 }{-2} \qquad \qquad\qquad x_{2} = \dfrac{-8 }{-2}$


$\boxed{ x_{1} = -3 \qquad \qquad\qquad x_{2} = 4 }$

Answer 2

- x2 + x + 12 = 0

Δ = b2 - 4 . a . c

Δ = 1 - 4 . (-1) . 12

Δ= 1 + 48

Δ = 49

x = - b ±$\sqrt{delta}$

           2.a

x' = - 1 + 7 / 2 . (-1)

x' = -1 +7 / - 2

x' = 6 / - 2

x' = - 3

x" = - 1 - 7 / 2 . ( - 1)

x" = - 8 / - 2

x" = 4