Question

x-5x+6=0 e equação de 2 grau

Answer 1

Utilizando a formula resolutiva: Bhaskara =>  $x= \frac{-b+- \sqrt{Delta} }{2a}$; sendo o discriminante Delta (Δ) => Δ= b²-4ac.

Tendo x²-5x+6=0, sendo a= 1; b= -5; c= 6
Δ= (-5)²-4*1*6

Δ= 25-24 = 1

$x= \frac{-(-5)+- \sqrt{1} }{2*1}$

$x= \frac{5+-1}{2}$

$x^1= \frac{5+1}{2} = 3$

$x^2= \frac{5-1}{2} = 2$

Answer 2

x-5x+6=0
delta=b2-4ac
delta=5ao quadrado -4.1.6
delta= 25-24
delta=1


x= -b+ou - raiz de delta
---------------------------------
2a


x= -(-5)+ou-1
___________
2.1


x'=5+1 = 6 = 3
______ __
2 2

x"= 5-1 =4 = 2
______ __
2 2

S={(3,2)}