<jupyter_start><jupyter_text>10 de agosto de 2018   Realizando equaciones a la antiguaVamos a realizar la siguiente ecuacion  $ x^3 - 7x^2 + 8x + 0.35 $<jupyter_code>def y(x):
    return x**3 - 7*x**2 + 8*x + 0.35

print(y(1.37))<jupyter_output>0.7430530000000001<jupyter_text>Analitico vs Numérico (Metodo Grafico)$ ax^2 + bx + c = 0$Analitico:$y = \frac{-b \pm \sqrt[\leftroot{-2}\uproot{2}]{b^2 - 4ac}}{2a}$Grafico:Darle valor a $x$ y graficar$2x^2 - 5x + 3 = 0$<jupyter_code>import numpy as np
import matplotlib.pyplot as plt

def y(x):
    return 1-((20**2/9.81)*(3+x)/(3*x+((x**2)/2))**3)

xarray = np.linspace(1, 4,100)

yarray = np.zeros(100)

for i in range(len(xarray)):
    yarray[i] = y(xarray[i])

plt.plot(xarray, yarray)
plt.grid()
plt.show()<jupyter_output><empty_output>
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