Question

me pueden ayudar por favor

Answer 1

Resolver:

1) $20 x^{2} +3x-2=0$

Aplicando la formula general: 

$x1 = \frac{-b+ \sqrt{b^2-4ac} }{2a} = \frac{-3+ \sqrt{3^2-4.20.(-2)} }{2.20} = \frac{-3+ \sqrt{9+160} }{40} = \frac{-3+ \sqrt{169} }{40} = \frac{-3+13}{40}$
$= \frac{x}{y} \frac{10}{40}= \frac{1}{4} = 0,25$

$x2=\frac{-b- \sqrt{b^2-4ac} }{2a}= \frac{-3- \sqrt{3^2-4.20.(-2)} }{2.20} = [tex]\frac{-3- \sqrt{9+160} }{40} = \frac{-3- \sqrt{169} }{40} = \frac{-3-13}{40}$ $= \frac{x}{y} \frac{-16}{40}= \frac{-4}{10} = \frac{-2}{5}$[/tex] =$-0,40$

2)
$5z^2=17z+14 =0$

$x1= \frac{-(-17)+ \sqrt{(-17)^2-4.5.14} }{2.5} = \frac{17+ \sqrt{289-280} }{2.5} = \frac{26}{10} = \frac{13}{5} = 2,6$

$x2= \frac{-(-17)- \sqrt{(-17)^2-4.5.14} }{2.5} = \frac{17- \sqrt{289-280} }{2.5} = \frac{14}{10} = \frac{7}{5} = 1,4$

3) 
10w^2-7w-6 = 0 

$x1= \frac{-(-7)+ \sqrt{(-7)^2-4.10.(-6)} }{2.10} = \frac{7+ \sqrt{49+240} }{20} = \frac{24}{20} = \frac{6}{5} = 1,20$

$x2= \frac{-(-7)- \sqrt{(-7)^2-4.10.(-6)} }{2.10} = \frac{7- \sqrt{49+240} }{20} = \frac{-10}{20} =- \frac{1}{2} = -0,50$

4)
14 x^{2} +17x-6=0

$x1= \frac{-17+ \sqrt{(-17)^2-4.14.(-6)} }{2.14} = \frac{-17+ \sqrt{289+336} }{28} = \frac{8}{28} = 0,29$

$x2= \frac{-17- \sqrt{(-17)^2-4.14.(-6)} }{2.14} = \frac{-17- \sqrt{289+336} }{28} = -\frac{42}{28} = -1,50$