Question

(x-3) (x+4) + 8 = x alguém pode me ajudar ???

Answer 1

$(x-3) (x+4) + 8 =0 \\ \\ x^{2} + 4x - 3x - 12 + 8 = 0 \\ \\ \\ x^{2} + x -4 = 0$

Resolvendo por Bháskara:

$x = \frac{-b \pm \sqrt{b^2 -4*a*c}}{2*a}$

a=1, b=1, c=−4
Δ=b2−4ac
Δ=(1)2−4*(1)*(−4)
Δ=1+16
Δ=17

$x = \frac{-b \pm \sqrt{\triangle}}{2*a}$

$x = \frac{-1 \pm \sqrt{17}}{2*1} \\ \\ x = \frac{-1 \pm \sqrt{17}}{2} \\ \\ \\ x' = \frac{-1 + \sqrt{17}}{2} \\ \\ \\ x'' = \frac{-1 - \sqrt{17}}{2}$

S= {$\frac{-1 + \sqrt{17}}{2}, \frac{-1 - \sqrt{17}}{2}$}