Question

como se resolve essa equação?


${x}^{2} + 2x + 40 = 80$

Answer 1

x² + 2x + 40 = 80
x² + 2x + 40 - 80 = 0
x² + 2x - 40 = 0

Δ = b² - 4ac
Δ = 2² - 4.1.(-40)
Δ = 4 + 160
Δ = 164

x = (-b +- √Δ)/2a
x = (-2 +- √164)/2
x = [-2 +- √(41.4)]/2
x = (-2 +- 2√41)/2
x = -1 +- √41

x' = -1 + √41
x" = -1 - √41

Se considerarmos que √41 = 6,4 temos:

x' = - 1 + 6,4 = 5,4
x" = - 1 - 6,4 = - 7,4 

=)

Answer 2

$\displaystyle \mathsf{x^2 + 2x + 40 = 80}\\ \displaystyle \mathsf{x^2 + 2x + 40 - 80 = 0}\\ \displaystyle \mathsf{x^2 + 2x - 40 = 0}$
$\displaystyle \mathsf{a = 1; b = 2; c = -40}\\ \\ \displaystyle \mathsf{\Delta = b^2 - 4.a.c}\\ \displaystyle \mathsf{\Delta = (2)^2- 4.(1).(-40)}\\ \displaystyle \mathsf{\Delta = 4 + 160}\\ \displaystyle \mathsf{\Delta = 164}\\ \\ \displaystyle \mathsf{x = \frac{-b \: \pm \: \sqrt{\Delta}}{2a}}\\ \\ \displaystyle \mathsf{x = \frac{-2 \: \pm \: 2\sqrt{41}}{2}}\\ \\ \displaystyle \mathsf{x' = -1 + \sqrt{41}}\\ \\ \displaystyle \mathsf{x'' = -1 -\sqrt{41}}$

Resposta: $\displaystyle \mathsf{x = -1 + \sqrt{41} \:ou\: x = -1 - \sqrt{41}}$