Question

Encontre Sá raizes da equação : 4x^2 + x - 5 =0

Answer 1

$4x^{2} +x-5=0$

Δ = b² - 4ac

Δ = 1² - 4.4.(-5)

Δ = 1 + 80

Δ = 81

$x=\frac{-b+-\sqrt{delta} }{2a} \\ \\ x=\frac{-1+-\sqrt{81} }{2.4} \\ \\ x=\frac{-1+-9}{8} \\ \\ x'=\frac{-1+9}{4} =-\frac{8}{4} =-2\\ \\ x''=\frac{-1-9}{8} =-\frac{10}{8} =-\frac{5}{4}$

Answer 2

4x² + x - 5 = 0

Δ= b² - 4ac
Δ= 1² - 4 * 4 * (-5)
Δ= 1 + 80
Δ= 81

x =( -b ± √Δ)/2*a
x =( - 1 ± √81)/2*4
x =( -1 ± 9)/8

x' = (-1 + 9)/8
x' = 8/8
x' = 1

x" = (-1 - 9)/8
x" = -10/8
x"= -5/4

S = {1, -5/4}