Question

$x^{2}-x-20=0$ achar delta, bhaskara

Answer 1

Resposta: Olá!

X^2-x-20=0

∆=b^2-4.a.b

a=1 b=-1 c=-20

∆=(-1)^2-4.1.(-20)

∆=1+80

∆=81

X=-b+-√∆/2.a

X1=1+9 = 5

2

X2=1-9 = -4

2

S={5,-4}

Espero ter ajudado!

Explicação passo-a-passo:

Answer 2

Olá, boa tarde!

Vamos utilizar a fórmula de Bhaskara.

$x^2-x-20=0\\\\Coeficientes:\;a=1,\;b=-1\;e\;c=-20\\\\\Delta=b^2-4ac\\\Delta=(-1)^2-4\cdot1\cdot(-20)\\\Delta=1+80\\\Delta=81\\\\x=\dfrac{-b\pm\sqrt{\Delta}}{2a}\\\\x=\dfrac{-(-1)\pm\sqrt{81}}{2\cdot1}\\\\x=\dfrac{1\pm9}{2}\\\\x_1=\dfrac{1+9}{2}=\dfrac{10}{2}=5\\\\x_2=\dfrac{1-9}{2}=\dfrac{-8}{2}=-4$

Espero ter ajudado.
Bons estudos! :)